C程序设计(第五版)谭浩强实验报告四(附答案以及源程序分析)codeblocks

谭浩强c++程序设计课后答案娄警卫第一章1.5题#include <iostream> using namespace std; int main(){cout<<"This"<<"is"; cout<<"a"<<"C++"; cout<<"program."; return 0;1.6题#include <iostream> using namespace std; int main(){int a,b,c;a=10;b=23;c=a+b;cout<<"a+b="; cout<<c;cout<<endl;return 0;}1.7七题#include <iostream> using namespace std; int main(){int a,b,c;int f(int x,int y,int z); cin>>a>>b>>c;c=f(a,b,c);cout<<c<<endl; return 0;}int f(int x,int y,int z) {int m;if (x<y) m=x;else m=y;if (z<m) m=z;return(m);}1.8题#include <iostream>using namespace std;int main(){int a,b,c;cin>>a>>b;c=a+b;cout<<"a+b="<<a+b<<endl; return 0;}1.9题#include <iostream>using namespace std;int main(){int a,b,c;int add(int x,int y); cin>>a>>b;c=add(a,b);cout<<"a+b="<<c<<endl; return 0;}int add(int x,int y){int z;z=x+y;return(z);}第二章2.3题#include <iostream>using namespace std;int main(){char c1='a',c2='b',c3='c',c4='\101',c5='\116'; cout<<c1<<c2<<c3<<'\n';cout<<"\t\b"<<c4<<'\t'<<c5<<'\n';return 0;}2.4题#include <iostream>using namespace std;int main(){char c1='C',c2='+',c3='+';cout<<"I say: \""<<c1<<c2<<c3<<'\"';cout<<"\t\t"<<"He says: \"C++ is very interesting!\""<< '\n';return 0;}2.7题#include <iostream>using namespace std;int main(){int i,j,m,n;i=8;j=10;m=++i+j++;n=(++i)+(++j)+m;cout<<i<<'\t'<<j<<'\t'<<m<<'\t'<<n<<endl; return 0;}2.8题#include <iostream>using namespace std;int main(){char c1='C', c2='h', c3='i', c4='n', c5='a';c1+=4;c2+=4;c3+=4;c4+=4;c5+=4;cout<<"password is:"<<c1<<c2<<c3<<c4<<c5<<endl;return 0;}第三章3.2题#include <iostream>#include <iomanip>using namespace std;int main ( ){float h,r,l,s,sq,vq,vz;const float pi=3.1415926;cout<<"please enter r,h:";cin>>r>>h;l=2*pi*r;s=r*r*pi;sq=4*pi*r*r;vq=3.0/4.0*pi*r*r*r;vz=pi*r*r*h;cout<<setiosflags(ios::fixed)<<setiosflags(ios:: right)<<setprecision(2);cout<<"l= "<<setw(10)<<l<<endl;cout<<"s= "<<setw(10)<<s<<endl;cout<<"sq="<<setw(10)<<sq<<endl;cout<<"vq="<<setw(10)<<vq<<endl;cout<<"vz="<<setw(10)<<vz<<endl;return 0;}3.3题#include <iostream>using namespace std;int main (){float c,f;cout<<"请输入一个华氏温度:";cin>>f;c=(5.0/9.0)*(f-32); //注意5和9要用实型表示,否则5/9值为0cout<<"摄氏温度为:"<<c<<endl;return 0;};3.4题#include <iostream>using namespace std;int main ( ){char c1,c2;cout<<"请输入两个字符c1,c2:";c1=getchar(); //将输入的第一个字符赋给c1c2=getchar(); //将输入的第二个字符赋给c2cout<<"用putchar函数输出结果为:"; putchar(c1);putchar(c2);cout<<endl;cout<<"用cout语句输出结果为:";cout<<c1<<c2<<endl;return 0;}3.4题另一解#include <iostream>using namespace std;int main ( ){char c1,c2;cout<<"请输入两个字符c1,c2:";c1=getchar(); //将输入的第一个字符赋给c1c2=getchar(); //将输入的第二个字符赋给c2cout<<"用putchar函数输出结果为:"; putchar(c1);putchar(44);putchar(c2);cout<<endl;cout<<"用cout语句输出结果为:";cout<<c1<<","<<c2<<endl;return 0;}3.5题#include <iostream>using namespace std;int main ( ){char c1,c2;int i1,i2; //定义为整型cout<<"请输入两个整数i1,i2:";cin>>i1>>i2;c1=i1;c2=i2;cout<<"按字符输出结果为:"<<c1<<" , "<<c2<<endl;return 0;}3.8题#include <iostream>using namespace std;int main ( ){ int a=3,b=4,c=5,x,y;cout<<(a+b>c && b==c)<<endl;cout<<(a||b+c && b-c)<<endl;cout<<(!(a>b) && !c||1)<<endl;cout<<(!(x=a) && (y=b) && 0)<<endl;cout<<(!(a+b)+c-1 && b+c/2)<<endl; return 0;}3.9题include <iostream>using namespace std;int main ( ){int a,b,c;cout<<"please enter three integer numbers:";cin>>a>>b>>c;if(a<b)if(b<c)cout<<"max="<<c;elsecout<<"max="<<b;else if (a<c)cout<<"max="<<c;elsecout<<"max="<<a;cout<<endl;return 0;}3.9题另一解#include <iostream>using namespace std;int main ( ){int a,b,c,temp,max ;cout<<"please enter three integer numbers:";cin>>a>>b>>c;temp=(a>b)?a:b; /* 将a和b中的大者存入temp中*/max=(temp>c)?temp:c; /* 将a和b中的大者与c比较，最大者存入max*/cout<<"max="<<max<<endl;return 0;}3.10题#include <iostream>using namespace std;int main ( ){int x,y;cout<<"enter x:";cin>>x;if (x<1){y=x;cout<<"x="<<x<<", y=x="<<y;}else if (x<10) // 1≤x＜10{y=2*x-1;cout<<"x="<<x<<", y=2*x-1="<<y;}else// x≥10{y=3*x-11;cout<<"x="<<x<<",y=3*x-11="<<y;}cout<<endl;return 0;}3.11题#include <iostream>using namespace std; int main (){float score;char grade;cout<<"please enter score of student:"; cin>>score;while (score>100||score<0){cout<<"data error,enter data again.";cin>>score;}switch(int(score/10)){case 10:case 9: grade='A';break;case 8: grade='B';break;case 7: grade='C';break;case 6: grade='D';break;default:grade='E';}cout<<"score is "<<score<<", grade is "<<grade<<endl;return 0;}3.12题#include <iostream>using namespace std;int main (){long int num;intindiv,ten,hundred,thousand,ten_thousand,pla ce;/*分别代表个位,十位,百位,千位,万位和位数*/cout<<"enter an integer(0~99999):"; cin>>num;if (num>9999)place=5;else if (num>999)place=4;else if (num>99)place=3;else if (num>9)place=2;else place=1;cout<<"place="<<place<<endl;//计算各位数字ten_thousand=num/10000;thousand=(int)(num-ten_thousand*10000)/1 000;hundred=(int)(num-ten_thousand*10000-tho usand*1000)/100;ten=(int)(num-ten_thousand*10000-thousan d*1000-hundred*100)/10;indiv=(int)(num-ten_thousand*10000-thousa nd*1000-hundred*100-ten*10);cout<<"original order:";switch(place){case5:cout<<ten_thousand<<","<<thousand<<","< <hundred<<","<<ten<<","<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<ten<<hundred<<thousand<<ten _thousand<<endl;break;case4:cout<<thousand<<","<<hundred<<","<<ten <<","<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<ten<<hundred<<thousand<<en dl;break;case3:cout<<hundred<<","<<ten<<","<<indiv<<en dl;cout<<"reverse order:";cout<<indiv<<ten<<hundred<<endl;break;case 2:cout<<ten<<","<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<ten<<endl;break;case 1:cout<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<endl;break;}return 0;}3.13题#include <iostream>using namespace std;int main (){ long i; //i为利润floatbonus,bon1,bon2,bon4,bon6,bon10;bon1=100000*0.1; //利润为10万元时的奖金bon2=bon1+100000*0.075; //利润为20万元时的奖金bon4=bon2+100000*0.05; //利润为40万元时的奖金bon6=bon4+100000*0.03; //利润为60万元时的奖金bon10=bon6+400000*0.015; //利润为100万元时的奖金cout<<"enter i:";cin>>i;if (i<=100000)bonus=i*0.1;//利润在10万元以内按10%提成奖金else if (i<=200000)bonus=bon1+(i-100000)*0.075; //利润在10万元至20万时的奖金else if (i<=400000)bonus=bon2+(i-200000)*0.05; //利润在20万元至40万时的奖金else if (i<=600000)bonus=bon4+(i-400000)*0.03; //利润在40万元至60万时的奖金else if (i<=1000000)bonus=bon6+(i-600000)*0.015; //利润在60万元至100万时的奖金elsebonus=bon10+(i-1000000)*0.01; //利润在100万元以上时的奖金cout<<"bonus="<<bonus<<endl;return 0;}3.13题另一解#include <iostream>using namespace std;int main (){long i;float bonus,bon1,bon2,bon4,bon6,bon10; int c;bon1=100000*0.1;bon2=bon1+100000*0.075;bon4=bon2+200000*0.05;bon6=bon4+200000*0.03;bon10=bon6+400000*0.015;cout<<"enter i:";cin>>i;c=i/100000;if (c>10) c=10;switch(c){case 0: bonus=i*0.1; break;case 1: bonus=bon1+(i-100000)*0.075; break;case 2:case3:bonus=bon2+(i-200000)*0.05;break;case 4:case5:bonus=bon4+(i-400000)*0.03;break;case 6:case 7:case 8:case 9: bonus=bon6+(i-600000)*0.015; break;case 10: bonus=bon10+(i-1000000)*0.01;}cout<<"bonus="<<bonus<<endl;return 0;}3.14题#include <iostream>using namespace std;int main (){int t,a,b,c,d;cout<<"enter four numbers:";cin>>a>>b>>c>>d;cout<<"a="<<a<<", b="<<b<<", c="<<c<<",d="<<d<<endl;if (a>b){t=a;a=b;b=t;}if (a>c){t=a; a=c; c=t;}if (a>d){t=a; a=d; d=t;}if (b>c){t=b; b=c; c=t;}if (b>d){t=b; b=d; d=t;}if (c>d){t=c; c=d; d=t;}cout<<"the sorted sequence:"<<endl;cout<<a<<", "<<b<<", "<<c<<", "<<d<<endl; return 0;}3.15题#include <iostream>using namespace std;int main (){int p,r,n,m,temp;cout<<"please enter two positive integer numbers n,m:";cin>>n>>m;if (n<m){temp=n;n=m;m=temp; //把大数放在n中, 小数放在m中}p=n*m; //先将n和m的乘积保存在p中, 以便求最小公倍数时用while (m!=0) //求n和m的最大公约数{r=n%m;n=m;m=r;}cout<<"HCF="<<n<<endl;cout<<"LCD="<<p/n<<endl; // p是原来两个整数的乘积return 0;}3.16题#include <iostream>using namespace std;int main (){char c;int letters=0,space=0,digit=0,other=0;cout<<"enter one line::"<<endl;while((c=getchar())!='\n'){if (c>='a' && c<='z'||c>='A' && c<='Z')letters++;else if (c==' ')space++;else if (c>='0' && c<='9')digit++;elseother++;}cout<<"letter:"<<letters<<", space:"<<space<<", digit:"<<digit<<", other:"<<other<<endl;return 0;}3.17题#include <iostream>using namespace std;int main (){int a,n,i=1,sn=0,tn=0;cout<<"a,n=:";cin>>a>>n;while (i<=n){tn=tn+a; //赋值后的tn为i个a 组成数的值sn=sn+tn; //赋值后的sn为多项式前i项之和a=a*10;++i;}cout<<"a+aa+aaa+...="<<sn<<endl;return 0;}3.18题#include <iostream>using namespace std;int main (){float s=0,t=1;int n;for (n=1;n<=20;n++){t=t*n; // 求n!s=s+t; // 将各项累加}cout<<"1!+2!+...+20!="<<s<<endl;return 0;}3.19题#include <iostream>using namespace std;int main (){int i,j,k,n;cout<<"narcissus numbers are:"<<endl;for (n=100;n<1000;n++){i=n/100;j=n/10-i*10;k=n%10;if (n == i*i*i + j*j*j + k*k*k)cout<<n<<" ";}cout<<endl;return 0;}3.20题#include <iostream>using namespace std;int main(){const int m=1000; // 定义寻找范围int k1,k2,k3,k4,k5,k6,k7,k8,k9,k10;int i,a,n,s;for (a=2;a<=m;a++) // a是2~1000之间的整数，检查它是否为完数{n=0; // n用来累计a的因子的个数s=a; // s用来存放尚未求出的因子之和，开始时等于afor (i=1;i<a;i++) // 检查i是否为a 的因子if (a%i==0) // 如果i是a的因子{n++; // n加1，表示新找到一个因子s=s-i; // s减去已找到的因子，s的新值是尚未求出的因子之和switch(n) // 将找到的因子赋给k1,...,k10{case 1:k1=i; break; // 找出的笫1个因子赋给k1case 2:k2=i; break; // 找出的笫2个因子赋给k2case 3:k3=i; break; // 找出的笫3个因子赋给k3case 4:k4=i; break; // 找出的笫4个因子赋给k4case 5:k5=i; break; // 找出的笫5个因子赋给k5case 6:k6=i; break; // 找出的笫6个因子赋给k6case 7:k7=i; break; // 找出的笫7个因子赋给k7case 8:k8=i; break; // 找出的笫8个因子赋给k8case 9:k9=i; break; // 找出的笫9个因子赋给k9case 10:k10=i; break; // 找出的笫10个因子赋给k10}}if (s==0) // s=0表示全部因子都已找到了{cout<<a<<" is a 完数"<<endl;cout<<"its factors are:";if (n>1) cout<<k1<<","<<k2; // n>1表示a至少有2个因子if (n>2) cout<<","<<k3; // n>2表示至少有3个因子，故应再输出一个因子if (n>3) cout<<","<<k4; // n>3表示至少有4个因子，故应再输出一个因子if (n>4) cout<<","<<k5; // 以下类似if (n>5) cout<<","<<k6;if (n>6) cout<<","<<k7;if (n>7) cout<<","<<k8;if (n>8) cout<<","<<k9;if (n>9) cout<<","<<k10;cout<<endl<<endl;}}return 0;}3.20题另一解#include <iostream>using namespace std;int main(){int m,s,i;for (m=2;m<1000;m++){s=0;for (i=1;i<m;i++)if ((m%i)==0) s=s+i;if(s==m){cout<<m<<" is a完数"<<endl;cout<<"its factors are:";for (i=1;i<m;i++)if (m%i==0) cout<<i<<" ";cout<<endl;}}return 0;}3.20题另一解#include <iostream>using namespace std;int main(){int k[11];int i,a,n,s;for (a=2;a<=1000;a++){n=0;s=a;for (i=1;i<a;i++)if ((a%i)==0){n++;s=s-i;k[n]=i; // 将找到的因子赋给k[1]┅k[10]}if (s==0){cout<<a<<" is a 完数"<<endl;cout<<"its factors are:";for (i=1;i<n;i++)cout<<k[i]<<" ";cout<<k[n]<<endl;}}return 0;}3.21题#include <iostream>using namespace std;int main(){int i,t,n=20;double a=2,b=1,s=0;for (i=1;i<=n;i++){s=s+a/b;t=a;a=a+b; // 将前一项分子与分母之和作为下一项的分子b=t; // 将前一项的分子作为下一项的分母}cout<<"sum="<<s<<endl;return 0;} 3.22题#include <iostream>using namespace std;int main(){int day,x1,x2;day=9;x2=1;while(day>0){x1=(x2+1)*2; // 第1天的桃子数是第2天桃子数加1后的2倍x2=x1;day--;}cout<<"total="<<x1<<endl;return 0;}3.23题#include <iostream>#include <cmath>using namespace std;int main(){float a,x0,x1;cout<<"enter a positive number:"; cin>>a; // 输入a的值x0=a/2;x1=(x0+a/x0)/2;do{x0=x1;x1=(x0+a/x0)/2;}while(fabs(x0-x1)>=1e-5);cout<<"The square root of "<<a<<" is "<<x1<<endl;return 0;}3.24题#include <iostream>using namespace std;int main(){int i,k;for (i=0;i<=3;i++) // 输出上面4行*号{for (k=0;k<=2*i;k++)cout<<"*"; // 输出*号cout<<endl; //输出完一行*号后换行}for (i=0;i<=2;i++) // 输出下面3行*号{for (k=0;k<=4-2*i;k++)cout<<"*"; // 输出*号cout<<endl; // 输出完一行*号后换行}return 0;}3.25题#include <iostream>using namespace std;int main(){char i,j,k; /* i是a的对手;j是b的对手;k是c的对手*/for (i='X';i<='Z';i++)for (j='X';j<='Z';j++)if (i!=j)for (k='X';k<='Z';k++)if (i!=k && j!=k)if (i!='X' && k!='X' && k!='Z')cout<<"A--"<<i<<"B--"<<j<<" C--"<<k<<endl;return 0;}第四章4.1题#include <iostream>using namespace std;int main(){int hcf(int,int);int lcd(int,int,int);int u,v,h,l;cin>>u>>v;h=hcf(u,v);cout<<"H.C.F="<<h<<endl;l=lcd(u,v,h);cout<<"L.C.D="<<l<<endl;return 0;}int hcf(int u,int v){int t,r;if (v>u){t=u;u=v;v=t;}while ((r=u%v)!=0){u=v;v=r;}return(v);}int lcd(int u,int v,int h){return(u*v/h);}4.2题#include <iostream>#include <math.h>using namespace std;float x1,x2,disc,p,q;int main(){void greater_than_zero(float,float); void equal_to_zero(float,float);void smaller_than_zero(float,float); float a,b,c;cout<<"input a,b,c:";cin>>a>>b>>c;disc=b*b-4*a*c;cout<<"root:"<<endl;if (disc>0){greater_than_zero(a,b);cout<<"x1="<<x1<<",x2="<<x2<<endl; }else if (disc==0){equal_to_zero(a,b);cout<<"x1="<<x1<<",x2="<<x2<<endl;}else{smaller_than_zero(a,b);cout<<"x1="<<p<<"+"<<q<<"i"<<endl;cout<<"x2="<<p<<"-"<<q<<"i"<<endl;}return 0;}void greater_than_zero(float a,float b) /* 定义一个函数，用来求disc>0时方程的根*/{x1=(-b+sqrt(disc))/(2*a);x2=(-b-sqrt(disc))/(2*a);}void equal_to_zero(float a,float b) /* 定义一个函数，用来求disc=0时方程的根*/{x1=x2=(-b)/(2*a);}void smaller_than_zero(float a,float b) /* 定义一个函数，用来求disc<0时方程的根*/{p=-b/(2*a);q=sqrt(-disc)/(2*a);}4.3题#include <iostream>using namespace std;int main(){int prime(int); /* 函数原型声明*/int n;cout<<"input an integer:";cin>>n;if (prime(n))cout<<n<<" is a prime."<<endl;elsecout<<n<<" is not a prime."<<endl;return 0;}int prime(int n){int flag=1,i;for (i=2;i<n/2 && flag==1;i++)if (n%i==0)flag=0;return(flag);}4.4题#include <iostream>using namespace std;int main(){int fac(int);int a,b,c,sum=0;cout<<"enter a,b,c:";cin>>a>>b>>c;sum=sum+fac(a)+fac(b)+fac(c);cout<<a<<"!+"<<b<<"!+"<<c<<"!="<<sum<<e ndl;return 0;}int fac(int n){int f=1;for (int i=1;i<=n;i++)f=f*i;return f;}4.5题#include <iostream>#include <cmath>using namespace std;int main(){double e(double);double x,sinh;cout<<"enter x:";cin>>x;sinh=(e(x)+e(-x))/2;cout<<"sinh("<<x<<")="<<sinh<<endl;return 0;}double e(double x){return exp(x);}4.6题#include <iostream>#include <cmath>using namespace std;int main(){doublesolut(double ,double ,double ,double ); double a,b,c,d;cout<<"input a,b,c,d:";cin>>a>>b>>c>>d;cout<<"x="<<solut(a,b,c,d)<<endl;return 0;}double solut(double a,double b,double c,double d){double x=1,x0,f,f1;do{x0=x;f=((a*x0+b)*x0+c)*x0+d;f1=(3*a*x0+2*b)*x0+c;x=x0-f/f1;}while(fabs(x-x0)>=1e-5);return(x);}4.7题#include <iostream>#include <cmath>using namespace std;int main(){void godbaha(int);int n;cout<<"input n:";cin>>n;godbaha(n);return 0;}void godbaha(int n){int prime(int);int a,b;for(a=3;a<=n/2;a=a+2){if(prime(a)){b=n-a;if (prime(b))cout<<n<<"="<<a<<"+"<<b<<endl;}}}int prime(int m){int i,k=sqrt(m);for(i=2;i<=k;i++)if(m%i==0) break;if (i>k) return 1;else return 0;}4.8题#include <iostream>using namespace std;int main(){int x,n;float p(int,int);cout<<"input n & x:";cin>>n>>x;cout<<"n="<<n<<",x="<<x<<endl;;cout<<"P"<<n<<"(x)="<<p(n,x)<<endl; return 0;}float p(int n,int x){if (n==0)return(1);else if (n==1)return(x);elsereturn(((2*n-1)*x*p((n-1),x)-(n-1)*p((n-2),x))/n);}4.9题#include <iostream>using namespace std;int main(){void hanoi(int n,char one,char two,char three);int m;cout<<"input the number of diskes:"; cin>>m;cout<<"The steps of moving "<<m<<" disks:"<<endl;hanoi(m,'A','B','C');return 0;}void hanoi(int n,char one,char two,char three) //将n个盘从one座借助two座，移到three座{void move(char x,char y);if(n==1) move(one,three);else{hanoi(n-1,one,three,two);move(one,three);hanoi(n-1,two,one,three);}}void move(char x,char y){cout<<x<<"-->"<<y<<endl;}4.10题#include <iostream>using namespace std;int main(){void convert(int n);int number;cout<<"input an integer:";cin>>number;cout<<"output:"<<endl;if (number<0){cout<<"-";number=-number;}convert(number);cout<<endl;return 0;}void convert(int n){int i;char c;if ((i=n/10)!=0)convert(i);c=n%10+'0';cout<<" "<<c;}4.11题#include <iostream>using namespace std;int main(){int f(int);int n,s;cout<<"input the number n:";cin>>n;s=f(n);cout<<"The result is "<<s<<endl;return 0;}int f(int n){;if (n==1)return 1;elsereturn (n*n+f(n-1));}4.12题#include <iostream>#include <cmath>using namespace std;#define S(a,b,c) (a+b+c)/2#define AREA(a,b,c)sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(S(a,b,c) -c))int main(){float a,b,c;cout<<"input a,b,c:";cin>>a>>b>>c;if (a+b>c && a+c>b && b+c>a)cout<<"area="<<AREA(a,b,c)<<endl; elsecout<<"It is not a triangle!"<<endl; return 0;}4.14题#include <iostream>using namespace std;//#define LETTER 1int main(){char c;cin>>c;#if LETTERif(c>='a' && c<='z')c=c-32;#elseif(c>='A' && c<='Z')c=c+32;#endifcout<<c<<endl;return 0;}4.15题#include <iostream>using namespace std;#define CHANGE 1int main(){char ch[40];cout<<"input text:"<<endl;;gets(ch);#if (CHANGE){for (int i=0;i<40;i++){if (ch[i]!='\0')if (ch[i]>='a'&& ch[i]<'z'||ch[i]>'A'&& ch[i]<'Z')ch[i]+=1;else if (ch[i]=='z'||ch[i]=='Z')ch[i]-=25;}}#endifcout<<"output:"<<endl<<ch<<endl;return 0;}4.16题file#include <iostream>using namespace std;int a;int main(){extern int power(int);int b=3,c,d,m;cout<<"enter an integer a and its power m:"<<endl;cin>>a>>m;c=a*b;cout<<a<<"*"<<b<<"="<<c<<endl;d=power(m);cout<<a<<"**"<<m<<"="<<d<<endl; return 0;}4.16题fileextern int a;int power(int n){int i,y=1;for(i=1;i<=n;i++)y*=a;return y;}第五章5.1题#include <iostream>#include <iomanip>using namespace std;#include <math.h>int main(){int i,j,n,a[101];for (i=1;i<=100;i++)a[i]=i;a[1]=0;for (i=2;i<sqrt(100);i++)for (j=i+1;j<=100;j++){if(a[i]!=0 && a[j]!=0)if (a[j]%a[i]==0)a[j]=0; }cout<<endl;for (i=1,n=0;i<=100;i++){if (a[i]!=0){cout<<setw(5)<<a[i]<<" ";n++;}if(n==10){cout<<endl;n=0;}}cout<<endl;return 0;}5.2题#include <iostream>using namespace std;//#include <math.h>int main(){int i,j,min,temp,a[11];cout<<"enter data:"<<endl;for (i=1;i<=10;i++){cout<<"a["<<i<<"]=";cin>>a[i]; //输入10个数}cout<<endl<<"The original numbers:"<<endl;;for (i=1;i<=10;i++)cout<<a[i]<<" "; // 输出这10个数cout<<endl;;for (i=1;i<=9;i++) //以下8行是对10个数排序{min=i;for (j=i+1;j<=10;j++)if (a[min]>a[j]) min=j;temp=a[i]; //以下3行将a[i+1]～a[10]中最小者与a[i] 对换a[i]=a[min];a[min]=temp;}cout<<endl<<"The sorted numbers:"<<endl;for (i=1;i<=10;i++) // 输出已排好序的10个数cout<<a[i]<<" ";cout<<endl;return 0;}5.3题#include <iostream>using namespace std;int main(){int a[3][3],sum=0;int i,j;cout<<"enter data:"<<endl;;for (i=0;i<3;i++)for (j=0;j<3;j++)cin>>a[i][j];for (i=0;i<3;i++)sum=sum+a[i][i];cout<<"sum="<<sum<<endl;return 0;}5.4题#include <iostream>using namespace std;int main(){int a[11]={1,4,6,9,13,16,19,28,40,100};int num,i,j;cout<<"array a:"<<endl;for (i=0;i<10;i++)cout<<a[i]<<" ";cout<<endl;;cout<<"insert data:";cin>>num;if (num>a[9])a[10]=num;else。

第一章程序设计和C语言【第15页】1-5#include <stdio.h>int main ( ){ printf ("**************************\n\n");printf(" Very Good!\n\n");printf ("**************************\n");return 0;}1-6#include <stdio.h>int main(){int a,b,c,max;printf("please input a,b,c:\n");scanf("%d,%d,%d",&a,&b,&c);max=a;if (max<b)max=b;if (max<c)max=c;printf("The largest number is %d\n",max);return 0;}第2章算法——程序的灵魂【第36页】暂无答案第3章最简单的C程序设计——顺序程序设计【第82页】3-1#include <stdio.h>#include <math.h>int main(){float p,r,n;r=0.1;n=10;p=pow(1+r,n);printf("p=%f\n",p);return 0;}3-2-1#include <stdio.h>#include <math.h>int main(){float r5,r3,r2,r1,r0,p,p1,p2,p3,p4,p5;p=1000;r5=0.0585;r2=0.0468;r1=0.0414;r0=0.0072;p1=p*((1+r5)*5); // 一次存5年期p2=p*(1+2*r2)*(1+3*r3); // 先存2年期，到期后将本息再存3年期p3=p*(1+3*r3)*(1+2*r2); // 先存3年期，到期后将本息再存2年期p4=p*pow(1+r1,5); // 存1年期，到期后将本息存再存1年期，连续存5次p5=p*pow(1+r0/4,4*5); // 存活期存款。

C语言程序设计实验报告学校：内蒙古科技大学经济与管理学院班级：信息管理与信息系统2011级姓名：王*学号：**********目录实验一C语言开发环境建立 (3)实验二C语言程序初步 (3)实验三变量与赋值 (4)实验四运算符与表达式 (5)实验五数据类型 (7)实验六控制台输入/输出 (9)实验七关系运算与逻辑运算 (11)实验八分支程序的测试与调试 (12)实验九条件型循环结构及其测试 (16)实验十计数型循环结构及其测试 (20)实验十一函数及其测试 (24)实验十二变量的储存性质 (26)实验十三数组 (30)实验十四指针 (32)实验十五结构体和共用体 (37)实验十六文件 (40)实验11、实验目的与要求（1）掌握C语言开发环境的软/硬件配置和安装方法。

（2）为了自己的方便进行工作环境配置。

（3）测试所建立的C语言程序开发环境。

2、实验测试一个简单的程序，对所建立的程序开发环境进行测试例如，可以准备一个简单的程序如下。

#include <stdio.h>int main(void){printf("I am studying programming with C.");}编译结果运行结果3.分析与预期结果相同实验21.实验目的与要求（1）进一步熟悉c语言程序设计环境。

（2）明确c语言程序的结构。

2. 程序检测与调试程序一错误程序：main()(printf("This is c program.");)调试结果：程序缺少“#include<stdio.h>”，main函数下面应该是“｛｝”而不是“（）”。

改正后的程序：#include<stdio.h>main(){printf("This is c program.\n");}程序二错误程序：#include<stdio.h>;main(){a=3,b=5;c=a*b;printf("The product is:",c)}调试结果：“#include<stdio.h>”后面没有“；”，对变量a,b,c没有定义，“｝”前缺少“；”。

C程序设计(第四版)(谭浩强)第五章课后习题答案循环结构程序设计P115 5.1 用while计算1至100的合.#include<stdio.h>int main(){int i=1,sum=0;while(i<=100) //对于需要运算的值,要么在运算前可以赋值,要么一开始要指定.{sum=sum+i;i++;}printf("The sum is %d .\n",sum);return 0;}P117 5.2 用do-while来做1至100的合.#include<stdio.h>int main(){int i=1,sum=0;do// do-while可以做的事,用while都可以做到.{ //do-while先做一次执行,再判断条件,而while却是先做一个条件,再执行.sum=sum+i;i++;}while(i<=100);printf("The sum is %d .\n",sum);return 0;}#include<stdio.h>int main(){int i,sum=0;printf("Please input a number :");scanf("%d",&i); //输入10以内,正常,11的话,则sum仍然是0.while(i<=10){sum=sum+i;i++;}printf("The sum of 1-10 is %d .\n",sum);return 0;}#include<stdio.h>int main(){int i,sum=0;printf("Please input a number :");scanf("%d",&i); //输入10以内,结果一样.输入11的话,先做操作,所以sum=11.do{sum=sum+i;i++;}while(i<=10); //此重点在于理解二者的差别.printf("The sum of 1-10 is %d .\n",sum);return 0;}#include<stdio.h>int main(){int i,b=0,a,c;for(i=0;i<=1000;i++){printf("Please input amount :"); //循环体内套有输出语句以及跳出语句.scanf("%d",&a);b=b+a;if(b>=100){break; //break是用于跳出循环,对if无效,对while for switch 这一类.}}c=b/i;printf("conut is %d , aver is %d ",i+1,c); //注意%号后的形式,否则可能输出错误.return 0;}P127 5.5 continue的例子.#include<stdio.h>int main(){int i;for(i=1;i<20;i++){if(i%3!=0){continue; //跳过本次I,执行下一个i.}printf("%d ",i);}printf("\n");return 0;}P128 5.6 形成一个4*5的矩阵.#include<stdio.h>int main(){int i,j,a=0; //没有给初值,会出现警告: 使用了未初始化的局部变量“a”.for(i=1;i<=4;i++){for(j=1;j<=5;j++,a++) // a用来控制换行.{if(a%5==0){printf("\n");}printf("%d\t",i*j);}}printf("\n");return 0;}P131 5.7 用一个交错的式子求哌的近似值.#include<stdio.h>#include<math.h>int main() //四分之哌等于(1)-(1/3)+(1/5)-(1/7)+(1/9)-(1/11).{float s=1,n=1,m,sum=0,t;for(m=1;;m=m+2) //不确定哪项才会小于等于十的负六次方,所以不指定,无限下去.{ //不指定第二项,执行语句中应该有控制跳出的句子,否则死循环.t=(s)*(n/m); //这是第一项,s是符号,if(fabs(t)<=1e-6) //应该写在这里,题目要求这一项不累加进去.{break;}sum=sum+t; //因为累加项在这里,所以,一旦跳出就不会累加进来了.s=s*(-1); //变号一次.}printf("四分之一哌的值是%f.\n",sum);printf("一个完整哌的值是%f.\n",sum*4);return 0;}//下面这段小代码用来验证关于数值型数据类型的关系.去掉注释,可运行.//如果出现类似值为全1或是全零的话,一般可以考虑数据类型赋值或是定义错了的问题.//#include <stdio.h> //这是关于int float double三者关复杂关系的,乱啊,如果看不懂,可以通过实验自己明白来.////int main() //在C语言中,1.0认为是实数,即是double型,所以,如果你把它用float输出的话,会有警告:警告 1 warning C4305: “=”: 从“double”到“float”截断.//{// float m,a,b,c; //一旦定义了是这种类型的话,输出或是赋值的时候只能扩展不能截断,意思就是能变成double型,不能变成int型啦.而且后面的赋值会跟着它变成相应的类型.比如下面的m=1,其实得到的是m=1.0.// int d,e,f;// m=1;// a=1.0/3;// b=1/3;// c=m/3;// d=1.0;// e=1/4;// f=1.0/4;// printf("%lf(float用double的%lf来输出是可以的.)\n%f\n%f\n%f\n",m,a,b,c); //不管上面定义什么,这边写的输出类型是什么,就按相应的类型输出,有可能会出错,所以建议按定义的类型来输出.当然扩展的是不会错的,截断的是会错的,比如float可以用%lf来输出,而不能用%d来输出.// printf("%d\n%d\n%d\n",d,e,f); //但是,不相应的int型不可以用%f来输出的.因为int float就不同种类,一个是整数,一个是小数,float double同样是有小数点的// return 0;//}#include<stdio.h>int main() //这个就是著名的Fibonacci(费波那契数列问题){int f1=1,f2=1,f3,i;printf("%12d\n%12d\n",f1,f2);for(i=1;i<=38;i++) //注意,这是个基础问题,(i=1;i<=5;i++)这里其实进行了次运算,因为有f1,f2,要求有个,所以要有个.要么写<=38,要么写<39,边界问题一定要注意,不可以太随意{f3=f1+f2;printf("%12d\n",f3); //这个问题同样适用于"一对兔子一个月生一对"的问题.,f1=f2; //f1=f1+f2;此时它们各是,所以,现在的f1是.f2=f3; //f2=f2+f1;此时的f1已经是最先二者之和了.可以不用到f3.}return 0;}#include<stdio.h>#include<math.h>int main(){double num;int i;printf("Please input a number :");scanf("%lf",&num); //因为sqrt要求是浮点型,那就给它浮点型,需要时再强制转换.for(i=2;i<=sqrt(num);i++) //这边是<=号没错.{if((int)num%i==0) //如果在这期间有任何一个可以为零的话,则不是素数.{break; //当然跳出.}} //执行到这里的时候,i=5,已经变成了!!if(i<=sqrt(num)){printf("Not %d",(int)num);}else//如上所述,i=5,超出了求根的值,所以是素数.{printf("Yes %d",(int)num);}return 0;}P137 5.10 求100至200间的素数.#include<stdio.h>//不解释,HOHO>>>>...#include<math.h>int main(){double j;int i,k=0;for(j=100;j<=200;j++){for(i=2;i<=sqrt(j);i++){if((int)j%i==0){break;}}k=k+1; //这里是布局的开头.学习一下,有助逻辑.if(i<=sqrt(j)){printf("Not %d ",(int)j);if(k%5==0) //5个换一次行.{printf("\n");}}else{printf("Yes %d ",(int)j);if(k%5==0){printf("\n");}}}return 0;}P139 5.11 密码转换.#include<stdio.h>int main(){char c;c=getchar();while(c!='\n') //这也可以用数组来实现.{if((c>='a'&&c<='z')||(c>='A'&&c<='Z')){if((c>='w'&&c<='z')||(c>='W'&&c<='Z')){c=c-22;}else{c=c+4;}printf("%c",c);c=getchar(); //套在循环里,依次得到字母,而while中判断回车为结束.}}printf("\n"); //这是布局问题.return 0;}P140 0.3 最大公约数和最小公倍数.#include<stdio.h>//最大公约数用累除法,除到无余数时的被除数是最大公约数.main (){int m, n, c, d;int gcd(); //这是最大公约数的缩写,此处调用函数,可以不写里面的实参.int lcm(); //这是最小公倍数的缩写,此处调用函数,可以不写里面的实参.printf("Please input two number :\n");scanf("%d %d",&m,&n);c=gcd(m,n); //c获取最大公约数d=lcm(m,n); //d获取最小公倍数printf("The GCD of %d and %d is : %d !\n", m, n, c);printf("The LCM of %d and %d is : %d !\n", m, n, d);return 0;}int gcd(int x, int y) //最大公约数Greatest Common Divisor{int temp;while(x%y!=0){temp=y; //y在下一轮中作为除数,即是下一轮中的X,所以先闪一边去.y=x%y; //x,y的余数作为下一轮中的Y,由x%y来取得.x=temp; //刚才temp中存储了y的值,现在拿出来作为下一轮中的X使用.}return y; //这是每一轮中的被除数,按原理来,这就是最大公约数,即累除法的原理. }int lcm(int x, int y) //最小公倍数Lowest Common Multiple{int i, temp;if(x<y) //此段代码结果是保证二者大的数在X上,小的数在Y上.即小于号降序.{ //以下为经典三行码,实现两个数的互换.temp=x;x=y;y=temp;}for(i=1; i<=y; i++) //设定一个区间,从1至大的数之间的循环.{if(!((x*i)%y)) //此式子如有余数,加上"!"号,会是假,则不返回,进行下一轮.{ //如此往复,直到取模无余数,那么小的数X乘以区间当前的I值,就是最小公倍数.return x*i;}}}P140 0.4 判断一串输入的字符.#include<stdio.h>int main(){char ch;int a=0,b=0,c=0,d=0,e=0;printf("Please input the string\n");while((ch=getchar())!='\n') //直到回车.{if(ch<='z'&&ch>='a'){a++;}else if(ch==' '){c++;}else if(ch<58&&ch>47){d++;}else if(ch<='Z'&&ch>='A'){b++;}else{e++;}}printf("大写%d 小写%d 空格%d 数字%d 其它%d\n",a,b,c,d,e); }#include<stdio.h>//不理解时可以百度或是谷歌更多的信息.int main() //想办法既快速做完,又要消化理解{int temp,i,a,n,sum=0; //主逻辑,友好性暂时放松.scanf("%d %d",&a,&n); //a是数字,n是要乘的个数.temp=a; //先把第一阶的值存起来.for(i=0;i<n;i++){sum=sum+a;printf("%d + ",a); //事关布局.a=a*10+temp; //重点是每次乘,然后加上上一个数.}printf("= %d .",sum);return 0;}P140 0.6 1!+2!+3!+4!.....的值.#include<stdio.h>int main() //1!+2!+3!+4!.....{int i,j,k,sum=0,m=1;scanf("%d",&k); //比如设定为,值为.for(i=1;i<=k;i++) //第一层循环,指定到.{for(j=1;j<=i;j++) //第二层循环,指定至当前数.{m=m*j;} //到此是阶乘的结构.sum=sum+m;m=1;}printf("%d",sum); //完全不理解时,搜索并参考.return 0; //尝试自己做,第一次做出来就是自己的东西了.}P140 0.7 1至100,1至50平方,1至10倒数的总和. #include<stdio.h>int main(){int a,b;double c,asum=0,bsum=0,csum=0;for(a=1;a<=100;a++) //三个块分别注释验证结果.{asum=asum+a;}for(b=1;b<=50;b++) //在VS运行中,注意*.cpp为C++语言.{bsum=bsum+b*b; //为了避免语言差别,请注意文件名为*.c.}for(c=1;c<=10;c++) //c作浮点运算,所以定义在double类型中.{csum=csum+1/c;}printf("%lf",asum+bsum+csum);return 0;}PP#include<stdio.h>#include<math.h>int main(){int j,k,s[6],x=100,y,sum=0;for(j=153;j<=154;j++){for(k=2;k>=0;k--){s[k]=j%(int)pow(10,k+1)/(int)pow(10,k);y=pow(s[k],3);sum+=y;printf("%d-%d--%d\t",k,s[k],j);printf("%d\n",sum);}return 0;}P140 0.8 水仙花数.//#include <stdio.h> //一步步的发现问题.////int main() //在%和/号之间,以前pow.以后再做.//{// int i,j,k,a,b,c,sum=0; //这里逻辑对,算出来却错了.// for(i=2;i<=4;i++) //计算机在想什么,看来它的大脑难以模拟.// {// for(j=pow(10,i);j<=pow(10,i+1)-1;j++) //我不完全明白它遵守的逻辑. // {// for(k=0;k<=i;k++)// {// sum+=pow((j%pow(10,i+1)/pow(10,i)),3);// }// if(sum==j)// {// printf("%d 是水仙花数!\n",j);// }// sum=0;// }// }// return 0;//} //为什么还是无法实现?!#include<stdio.h>#include<math.h>int main(){int a,b,c,i,sum=0; //这里只计算三位数的.for(i=100;i<1000;i++){a=i/100;b=i%100/10;c=i%10;sum+=a*a*a+b*b*b+c*c*c;if(sum==i)printf("%d 是水仙花数.\n",i);}sum=0;}return 0;}P141 0.9 完数.#include<stdio.h>int main(){int i,j,r;for(i=1;i<=1000;i++) //零是个临界值,不能包括它.{r=0; //每次清零重来.类似水仙中的sum.for(j=1;j<i;j++){if(i%j==0) //除得尽即是因子.{r=r+j; //然后累加进去.}}if(r==i) //若相等.{printf("%d 是完数.\n",i);}}return 0;}P141 0.10 2/1+3/2+5/3+8/5+13/8…#include<stdio.h>int main() //10.007051{double i,a=2,b=1,c,s=0; //a是分子,b是分母.for(i=0;i<6;i++) //二十可以用户指定.{s+=a/b; //中间储值变量.c=a+b;b=a;a=c;}printf("%lf\n",s);return 0;}#include<stdio.h>int main() //从一百开始,减一半再自加两次,下降一次,反弹一次.{double sum=100,high=100,up,donw,i;for(i=1;i<10;i++) //不管指定到哪个数,都不会超过三百.{up=high/2;donw=up;high=donw; //自咬尾巴问题.sum+=donw*2;}printf("%lf %lf\n",donw,sum);return 0;}P141 0.12 猴子吃桃子.#include<stdio.h>int main() //从一百开始,减一半再自加两次,下降一次,反弹一次.{int i,sum=1;for(i=1;i<=10;i++) //临界要清楚,结果要与手算的前几个实例相匹配.{printf("倒数第%d天还剩有%d个桃子.\n",i,sum);sum=(sum+1)*2;}//printf("%d\n",sum);return 0;}P141 0.16 输出星号棱形图案.#include<stdio.h>#include<conio.h>main() //只关注左半部分.右半部分无视空格.{int i,j,k,m=6; //m可指定,指定中心点位置.可任意奇偶.for(i=1;i<=m;i++) //上半部分.其实也是正三角.{for(j=1;j<=m-i;j++) //一到中心点前i个位置填充空格.printf(" ");for(k=1;k<2*i;k++) //空格后向前填充星号的个数.是奇数.一,三,五... printf("*");printf("\n");}for(i=m-1;i>0;i--) //下半部分.其实也是倒三角.{ //减一是因为行数问题.这是中心行以下的.for(j=m-1;j>=i;j--)printf(" ");for(k=1;k<2*i;k++)printf("*");printf("\n");}getch(); //用户反应后结束.但,没必要.}。

《C程序设计》实验报告年级专业班姓名成绩课程名称C语言实验项目名称循环结构指导教师一、实验目的1、熟练掌握while语句。

2、熟练掌握do---while语句。

3、熟练掌握for语句。

4、熟练掌握各种选择语句与循环语句的嵌套使用。

5、熟练掌握判断素数的算法。

6、熟练掌握累加等常见的算法。

二、实验内容4.1求s=1＋1/2＋1/3＋..＋1/n，输入n的值，输出s的值保留两位小数。

4.2输入一行字符，分别统计出其中英文字母、空格、数字和其他字符的个数。

4.3 输出1000以内的“水仙花数”，所谓“水仙花数”是指一个3位数，其各位数字立方和等于该数本身，例如：153是水仙花数，因为153=13+53+33。

4.4 一个球从100m高度自由落下，每次落地后反弹回原高度的一半，再落下，再反弹，求它在第10次落地时共经过多少米，第10次反弹多高。

4.5 判断素数，输入任意整数，如果是输出Y es，如果不是输出No。

4.6 求s=a+aa+aaa+………，例如：输入a的值5，n的值6，则：S=5+55+555+5555+55555+5555554.7 编写程序，打印以下图形：（提示：使用双重循环）6.#include<stdio.h>void main(){int i,j,a,n,Sn=0;scanf("%ld%ld",&a,&n);j=a;for(i=1;i<=n;i++){Sn=Sn+a;a=j+10*a;}printf("%ld+%ld%ld+%ld%ld%ld+…=%ld\n",j,j,j,j,j,j,Sn); }7.#include<stdio.h>int main(void){int i,j;for(i=1;i<=7;i+=2){for (j=0;j++<i;printf("*"));printf("\n");}for(i-=4;i>0;i-=2)return 0;}四、运行结果（将每道题的运行结果截图并粘贴在此处）1.3.4.5.6.7.五、实验结果分析六、1.第四行程序“double”浮点型输入，第五行程序“scanf”标准输入n“while”第六行循环体语句循环条件i<=n。

谭浩强c++程序设计课后答案娄警卫第一章1.5题#include <iostream> using namespace std; int main(){cout<<"This"<<"is"; cout<<"a"<<"C++"; cout<<"program."; return 0;1.6题#include <iostream> using namespace std; int main(){int a,b,c;a=10;b=23;c=a+b;cout<<"a+b="; cout<<c;cout<<endl;return 0;}1.7七题#include <iostream> using namespace std; int main(){int a,b,c;int f(int x,int y,int z); cin>>a>>b>>c;c=f(a,b,c);cout<<c<<endl; return 0;}int f(int x,int y,int z) {int m;if (x<y) m=x;else m=y;if (z<m) m=z;return(m);}1.8题#include <iostream>using namespace std;int main(){int a,b,c;cin>>a>>b;c=a+b;cout<<"a+b="<<a+b<<endl; return 0;}1.9题#include <iostream>using namespace std;int main(){int a,b,c;int add(int x,int y); cin>>a>>b;c=add(a,b);cout<<"a+b="<<c<<endl; return 0;}int add(int x,int y){int z;z=x+y;return(z);}第二章2.3题#include <iostream>using namespace std;int main(){char c1='a',c2='b',c3='c',c4='\101',c5='\116'; cout<<c1<<c2<<c3<<'\n';cout<<"\t\b"<<c4<<'\t'<<c5<<'\n';return 0;}2.4题#include <iostream>using namespace std;int main(){char c1='C',c2='+',c3='+';cout<<"I say: \""<<c1<<c2<<c3<<'\"';cout<<"\t\t"<<"He says: \"C++ is very interesting!\""<< '\n';return 0;}2.7题#include <iostream>using namespace std;int main(){int i,j,m,n;i=8;j=10;m=++i+j++;n=(++i)+(++j)+m;cout<<i<<'\t'<<j<<'\t'<<m<<'\t'<<n<<endl; return 0;}2.8题#include <iostream>using namespace std;int main(){char c1='C', c2='h', c3='i', c4='n', c5='a';c1+=4;c2+=4;c3+=4;c4+=4;c5+=4;cout<<"password is:"<<c1<<c2<<c3<<c4<<c5<<endl;return 0;}第三章3.2题#include <iostream>#include <iomanip>using namespace std;int main ( ){float h,r,l,s,sq,vq,vz;const float pi=3.1415926;cout<<"please enter r,h:";cin>>r>>h;l=2*pi*r;s=r*r*pi;sq=4*pi*r*r;vq=3.0/4.0*pi*r*r*r;vz=pi*r*r*h;cout<<setiosflags(ios::fixed)<<setiosflags(ios:: right)<<setprecision(2);cout<<"l= "<<setw(10)<<l<<endl;cout<<"s= "<<setw(10)<<s<<endl;cout<<"sq="<<setw(10)<<sq<<endl;cout<<"vq="<<setw(10)<<vq<<endl;cout<<"vz="<<setw(10)<<vz<<endl;return 0;}3.3题#include <iostream>using namespace std;int main (){float c,f;cout<<"请输入一个华氏温度:";cin>>f;c=(5.0/9.0)*(f-32); //注意5和9要用实型表示,否则5/9值为0cout<<"摄氏温度为:"<<c<<endl;return 0;};3.4题#include <iostream>using namespace std;int main ( ){char c1,c2;cout<<"请输入两个字符c1,c2:";c1=getchar(); //将输入的第一个字符赋给c1c2=getchar(); //将输入的第二个字符赋给c2cout<<"用putchar函数输出结果为:"; putchar(c1);putchar(c2);cout<<endl;cout<<"用cout语句输出结果为:";cout<<c1<<c2<<endl;return 0;}3.4题另一解#include <iostream>using namespace std;int main ( ){char c1,c2;cout<<"请输入两个字符c1,c2:";c1=getchar(); //将输入的第一个字符赋给c1c2=getchar(); //将输入的第二个字符赋给c2cout<<"用putchar函数输出结果为:"; putchar(c1);putchar(44);putchar(c2);cout<<endl;cout<<"用cout语句输出结果为:";cout<<c1<<","<<c2<<endl;return 0;}3.5题#include <iostream>using namespace std;int main ( ){char c1,c2;int i1,i2; //定义为整型cout<<"请输入两个整数i1,i2:";cin>>i1>>i2;c1=i1;c2=i2;cout<<"按字符输出结果为:"<<c1<<" , "<<c2<<endl;return 0;}3.8题#include <iostream>using namespace std;int main ( ){ int a=3,b=4,c=5,x,y;cout<<(a+b>c && b==c)<<endl;cout<<(a||b+c && b-c)<<endl;cout<<(!(a>b) && !c||1)<<endl;cout<<(!(x=a) && (y=b) && 0)<<endl;cout<<(!(a+b)+c-1 && b+c/2)<<endl; return 0;}3.9题include <iostream>using namespace std;int main ( ){int a,b,c;cout<<"please enter three integer numbers:";cin>>a>>b>>c;if(a<b)if(b<c)cout<<"max="<<c;elsecout<<"max="<<b;else if (a<c)cout<<"max="<<c;elsecout<<"max="<<a;cout<<endl;return 0;}3.9题另一解#include <iostream>using namespace std;int main ( ){int a,b,c,temp,max ;cout<<"please enter three integer numbers:";cin>>a>>b>>c;temp=(a>b)?a:b; /* 将a和b中的大者存入temp中*/max=(temp>c)?temp:c; /* 将a和b中的大者与c比较，最大者存入max*/cout<<"max="<<max<<endl;return 0;}3.10题#include <iostream>using namespace std;int main ( ){int x,y;cout<<"enter x:";cin>>x;if (x<1){y=x;cout<<"x="<<x<<", y=x="<<y;}else if (x<10) // 1≤x＜10{y=2*x-1;cout<<"x="<<x<<", y=2*x-1="<<y;}else// x≥10{y=3*x-11;cout<<"x="<<x<<",y=3*x-11="<<y;}cout<<endl;return 0;}3.11题#include <iostream>using namespace std; int main (){float score;char grade;cout<<"please enter score of student:"; cin>>score;while (score>100||score<0){cout<<"data error,enter data again.";cin>>score;}switch(int(score/10)){case 10:case 9: grade='A';break;case 8: grade='B';break;case 7: grade='C';break;case 6: grade='D';break;default:grade='E';}cout<<"score is "<<score<<", grade is "<<grade<<endl;return 0;}3.12题#include <iostream>using namespace std;int main (){long int num;intindiv,ten,hundred,thousand,ten_thousand,pla ce;/*分别代表个位,十位,百位,千位,万位和位数*/cout<<"enter an integer(0~99999):"; cin>>num;if (num>9999)place=5;else if (num>999)place=4;else if (num>99)place=3;else if (num>9)place=2;else place=1;cout<<"place="<<place<<endl;//计算各位数字ten_thousand=num/10000;thousand=(int)(num-ten_thousand*10000)/1 000;hundred=(int)(num-ten_thousand*10000-tho usand*1000)/100;ten=(int)(num-ten_thousand*10000-thousan d*1000-hundred*100)/10;indiv=(int)(num-ten_thousand*10000-thousa nd*1000-hundred*100-ten*10);cout<<"original order:";switch(place){case5:cout<<ten_thousand<<","<<thousand<<","< <hundred<<","<<ten<<","<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<ten<<hundred<<thousand<<ten _thousand<<endl;break;case4:cout<<thousand<<","<<hundred<<","<<ten <<","<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<ten<<hundred<<thousand<<en dl;break;case3:cout<<hundred<<","<<ten<<","<<indiv<<en dl;cout<<"reverse order:";cout<<indiv<<ten<<hundred<<endl;break;case 2:cout<<ten<<","<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<ten<<endl;break;case 1:cout<<indiv<<endl;cout<<"reverse order:";cout<<indiv<<endl;break;}return 0;}3.13题#include <iostream>using namespace std;int main (){ long i; //i为利润floatbonus,bon1,bon2,bon4,bon6,bon10;bon1=100000*0.1; //利润为10万元时的奖金bon2=bon1+100000*0.075; //利润为20万元时的奖金bon4=bon2+100000*0.05; //利润为40万元时的奖金bon6=bon4+100000*0.03; //利润为60万元时的奖金bon10=bon6+400000*0.015; //利润为100万元时的奖金cout<<"enter i:";cin>>i;if (i<=100000)bonus=i*0.1;//利润在10万元以内按10%提成奖金else if (i<=200000)bonus=bon1+(i-100000)*0.075; //利润在10万元至20万时的奖金else if (i<=400000)bonus=bon2+(i-200000)*0.05; //利润在20万元至40万时的奖金else if (i<=600000)bonus=bon4+(i-400000)*0.03; //利润在40万元至60万时的奖金else if (i<=1000000)bonus=bon6+(i-600000)*0.015; //利润在60万元至100万时的奖金elsebonus=bon10+(i-1000000)*0.01; //利润在100万元以上时的奖金cout<<"bonus="<<bonus<<endl;return 0;}3.13题另一解#include <iostream>using namespace std;int main (){long i;float bonus,bon1,bon2,bon4,bon6,bon10; int c;bon1=100000*0.1;bon2=bon1+100000*0.075;bon4=bon2+200000*0.05;bon6=bon4+200000*0.03;bon10=bon6+400000*0.015;cout<<"enter i:";cin>>i;c=i/100000;if (c>10) c=10;switch(c){case 0: bonus=i*0.1; break;case 1: bonus=bon1+(i-100000)*0.075; break;case 2:case3:bonus=bon2+(i-200000)*0.05;break;case 4:case5:bonus=bon4+(i-400000)*0.03;break;case 6:case 7:case 8:case 9: bonus=bon6+(i-600000)*0.015; break;case 10: bonus=bon10+(i-1000000)*0.01;}cout<<"bonus="<<bonus<<endl;return 0;}3.14题#include <iostream>using namespace std;int main (){int t,a,b,c,d;cout<<"enter four numbers:";cin>>a>>b>>c>>d;cout<<"a="<<a<<", b="<<b<<", c="<<c<<",d="<<d<<endl;if (a>b){t=a;a=b;b=t;}if (a>c){t=a; a=c; c=t;}if (a>d){t=a; a=d; d=t;}if (b>c){t=b; b=c; c=t;}if (b>d){t=b; b=d; d=t;}if (c>d){t=c; c=d; d=t;}cout<<"the sorted sequence:"<<endl;cout<<a<<", "<<b<<", "<<c<<", "<<d<<endl; return 0;}3.15题#include <iostream>using namespace std;int main (){int p,r,n,m,temp;cout<<"please enter two positive integer numbers n,m:";cin>>n>>m;if (n<m){temp=n;n=m;m=temp; //把大数放在n中, 小数放在m中}p=n*m; //先将n和m的乘积保存在p中, 以便求最小公倍数时用while (m!=0) //求n和m的最大公约数{r=n%m;n=m;m=r;}cout<<"HCF="<<n<<endl;cout<<"LCD="<<p/n<<endl; // p是原来两个整数的乘积return 0;}3.16题#include <iostream>using namespace std;int main (){char c;int letters=0,space=0,digit=0,other=0;cout<<"enter one line::"<<endl;while((c=getchar())!='\n'){if (c>='a' && c<='z'||c>='A' && c<='Z')letters++;else if (c==' ')space++;else if (c>='0' && c<='9')digit++;elseother++;}cout<<"letter:"<<letters<<", space:"<<space<<", digit:"<<digit<<", other:"<<other<<endl;return 0;}3.17题#include <iostream>using namespace std;int main (){int a,n,i=1,sn=0,tn=0;cout<<"a,n=:";cin>>a>>n;while (i<=n){tn=tn+a; //赋值后的tn为i个a 组成数的值sn=sn+tn; //赋值后的sn为多项式前i项之和a=a*10;++i;}cout<<"a+aa+aaa+...="<<sn<<endl;return 0;}3.18题#include <iostream>using namespace std;int main (){float s=0,t=1;int n;for (n=1;n<=20;n++){t=t*n; // 求n!s=s+t; // 将各项累加}cout<<"1!+2!+...+20!="<<s<<endl;return 0;}3.19题#include <iostream>using namespace std;int main (){int i,j,k,n;cout<<"narcissus numbers are:"<<endl;for (n=100;n<1000;n++){i=n/100;j=n/10-i*10;k=n%10;if (n == i*i*i + j*j*j + k*k*k)cout<<n<<" ";}cout<<endl;return 0;}3.20题#include <iostream>using namespace std;int main(){const int m=1000; // 定义寻找范围int k1,k2,k3,k4,k5,k6,k7,k8,k9,k10;int i,a,n,s;for (a=2;a<=m;a++) // a是2~1000之间的整数，检查它是否为完数{n=0; // n用来累计a的因子的个数s=a; // s用来存放尚未求出的因子之和，开始时等于afor (i=1;i<a;i++) // 检查i是否为a 的因子if (a%i==0) // 如果i是a的因子{n++; // n加1，表示新找到一个因子s=s-i; // s减去已找到的因子，s的新值是尚未求出的因子之和switch(n) // 将找到的因子赋给k1,...,k10{case 1:k1=i; break; // 找出的笫1个因子赋给k1case 2:k2=i; break; // 找出的笫2个因子赋给k2case 3:k3=i; break; // 找出的笫3个因子赋给k3case 4:k4=i; break; // 找出的笫4个因子赋给k4case 5:k5=i; break; // 找出的笫5个因子赋给k5case 6:k6=i; break; // 找出的笫6个因子赋给k6case 7:k7=i; break; // 找出的笫7个因子赋给k7case 8:k8=i; break; // 找出的笫8个因子赋给k8case 9:k9=i; break; // 找出的笫9个因子赋给k9case 10:k10=i; break; // 找出的笫10个因子赋给k10}}if (s==0) // s=0表示全部因子都已找到了{cout<<a<<" is a 完数"<<endl;cout<<"its factors are:";if (n>1) cout<<k1<<","<<k2; // n>1表示a至少有2个因子if (n>2) cout<<","<<k3; // n>2表示至少有3个因子，故应再输出一个因子if (n>3) cout<<","<<k4; // n>3表示至少有4个因子，故应再输出一个因子if (n>4) cout<<","<<k5; // 以下类似if (n>5) cout<<","<<k6;if (n>6) cout<<","<<k7;if (n>7) cout<<","<<k8;if (n>8) cout<<","<<k9;if (n>9) cout<<","<<k10;cout<<endl<<endl;}}return 0;}3.20题另一解#include <iostream>using namespace std;int main(){int m,s,i;for (m=2;m<1000;m++){s=0;for (i=1;i<m;i++)if ((m%i)==0) s=s+i;if(s==m){cout<<m<<" is a完数"<<endl;cout<<"its factors are:";for (i=1;i<m;i++)if (m%i==0) cout<<i<<" ";cout<<endl;}}return 0;}3.20题另一解#include <iostream>using namespace std;int main(){int k[11];int i,a,n,s;for (a=2;a<=1000;a++){n=0;s=a;for (i=1;i<a;i++)if ((a%i)==0){n++;s=s-i;k[n]=i; // 将找到的因子赋给k[1]┅k[10]}if (s==0){cout<<a<<" is a 完数"<<endl;cout<<"its factors are:";for (i=1;i<n;i++)cout<<k[i]<<" ";cout<<k[n]<<endl;}}return 0;}3.21题#include <iostream>using namespace std;int main(){int i,t,n=20;double a=2,b=1,s=0;for (i=1;i<=n;i++){s=s+a/b;t=a;a=a+b; // 将前一项分子与分母之和作为下一项的分子b=t; // 将前一项的分子作为下一项的分母}cout<<"sum="<<s<<endl;return 0;} 3.22题#include <iostream>using namespace std;int main(){int day,x1,x2;day=9;x2=1;while(day>0){x1=(x2+1)*2; // 第1天的桃子数是第2天桃子数加1后的2倍x2=x1;day--;}cout<<"total="<<x1<<endl;return 0;}3.23题#include <iostream>#include <cmath>using namespace std;int main(){float a,x0,x1;cout<<"enter a positive number:"; cin>>a; // 输入a的值x0=a/2;x1=(x0+a/x0)/2;do{x0=x1;x1=(x0+a/x0)/2;}while(fabs(x0-x1)>=1e-5);cout<<"The square root of "<<a<<" is "<<x1<<endl;return 0;}3.24题#include <iostream>using namespace std;int main(){int i,k;for (i=0;i<=3;i++) // 输出上面4行*号{for (k=0;k<=2*i;k++)cout<<"*"; // 输出*号cout<<endl; //输出完一行*号后换行}for (i=0;i<=2;i++) // 输出下面3行*号{for (k=0;k<=4-2*i;k++)cout<<"*"; // 输出*号cout<<endl; // 输出完一行*号后换行}return 0;}3.25题#include <iostream>using namespace std;int main(){char i,j,k; /* i是a的对手;j是b的对手;k是c的对手*/for (i='X';i<='Z';i++)for (j='X';j<='Z';j++)if (i!=j)for (k='X';k<='Z';k++)if (i!=k && j!=k)if (i!='X' && k!='X' && k!='Z')cout<<"A--"<<i<<"B--"<<j<<" C--"<<k<<endl;return 0;}第四章4.1题#include <iostream>using namespace std;int main(){int hcf(int,int);int lcd(int,int,int);int u,v,h,l;cin>>u>>v;h=hcf(u,v);cout<<"H.C.F="<<h<<endl;l=lcd(u,v,h);cout<<"L.C.D="<<l<<endl;return 0;}int hcf(int u,int v){int t,r;if (v>u){t=u;u=v;v=t;}while ((r=u%v)!=0){u=v;v=r;}return(v);}int lcd(int u,int v,int h){return(u*v/h);}4.2题#include <iostream>#include <math.h>using namespace std;float x1,x2,disc,p,q;int main(){void greater_than_zero(float,float); void equal_to_zero(float,float);void smaller_than_zero(float,float); float a,b,c;cout<<"input a,b,c:";cin>>a>>b>>c;disc=b*b-4*a*c;cout<<"root:"<<endl;if (disc>0){greater_than_zero(a,b);cout<<"x1="<<x1<<",x2="<<x2<<endl; }else if (disc==0){equal_to_zero(a,b);cout<<"x1="<<x1<<",x2="<<x2<<endl;}else{smaller_than_zero(a,b);cout<<"x1="<<p<<"+"<<q<<"i"<<endl;cout<<"x2="<<p<<"-"<<q<<"i"<<endl;}return 0;}void greater_than_zero(float a,float b) /* 定义一个函数，用来求disc>0时方程的根*/{x1=(-b+sqrt(disc))/(2*a);x2=(-b-sqrt(disc))/(2*a);}void equal_to_zero(float a,float b) /* 定义一个函数，用来求disc=0时方程的根*/{x1=x2=(-b)/(2*a);}void smaller_than_zero(float a,float b) /* 定义一个函数，用来求disc<0时方程的根*/{p=-b/(2*a);q=sqrt(-disc)/(2*a);}4.3题#include <iostream>using namespace std;int main(){int prime(int); /* 函数原型声明*/int n;cout<<"input an integer:";cin>>n;if (prime(n))cout<<n<<" is a prime."<<endl;elsecout<<n<<" is not a prime."<<endl;return 0;}int prime(int n){int flag=1,i;for (i=2;i<n/2 && flag==1;i++)if (n%i==0)flag=0;return(flag);}4.4题#include <iostream>using namespace std;int main(){int fac(int);int a,b,c,sum=0;cout<<"enter a,b,c:";cin>>a>>b>>c;sum=sum+fac(a)+fac(b)+fac(c);cout<<a<<"!+"<<b<<"!+"<<c<<"!="<<sum<<e ndl;return 0;}int fac(int n){int f=1;for (int i=1;i<=n;i++)f=f*i;return f;}4.5题#include <iostream>#include <cmath>using namespace std;int main(){double e(double);double x,sinh;cout<<"enter x:";cin>>x;sinh=(e(x)+e(-x))/2;cout<<"sinh("<<x<<")="<<sinh<<endl;return 0;}double e(double x){return exp(x);}4.6题#include <iostream>#include <cmath>using namespace std;int main(){doublesolut(double ,double ,double ,double ); double a,b,c,d;cout<<"input a,b,c,d:";cin>>a>>b>>c>>d;cout<<"x="<<solut(a,b,c,d)<<endl;return 0;}double solut(double a,double b,double c,double d){double x=1,x0,f,f1;do{x0=x;f=((a*x0+b)*x0+c)*x0+d;f1=(3*a*x0+2*b)*x0+c;x=x0-f/f1;}while(fabs(x-x0)>=1e-5);return(x);}4.7题#include <iostream>#include <cmath>using namespace std;int main(){void godbaha(int);int n;cout<<"input n:";cin>>n;godbaha(n);return 0;}void godbaha(int n){int prime(int);int a,b;for(a=3;a<=n/2;a=a+2){if(prime(a)){b=n-a;if (prime(b))cout<<n<<"="<<a<<"+"<<b<<endl;}}}int prime(int m){int i,k=sqrt(m);for(i=2;i<=k;i++)if(m%i==0) break;if (i>k) return 1;else return 0;}4.8题#include <iostream>using namespace std;int main(){int x,n;float p(int,int);cout<<"input n & x:";cin>>n>>x;cout<<"n="<<n<<",x="<<x<<endl;;cout<<"P"<<n<<"(x)="<<p(n,x)<<endl; return 0;}float p(int n,int x){if (n==0)return(1);else if (n==1)return(x);elsereturn(((2*n-1)*x*p((n-1),x)-(n-1)*p((n-2),x))/n);}4.9题#include <iostream>using namespace std;int main(){void hanoi(int n,char one,char two,char three);int m;cout<<"input the number of diskes:"; cin>>m;cout<<"The steps of moving "<<m<<" disks:"<<endl;hanoi(m,'A','B','C');return 0;}void hanoi(int n,char one,char two,char three) //将n个盘从one座借助two座，移到three座{void move(char x,char y);if(n==1) move(one,three);else{hanoi(n-1,one,three,two);move(one,three);hanoi(n-1,two,one,three);}}void move(char x,char y){cout<<x<<"-->"<<y<<endl;}4.10题#include <iostream>using namespace std;int main(){void convert(int n);int number;cout<<"input an integer:";cin>>number;cout<<"output:"<<endl;if (number<0){cout<<"-";number=-number;}convert(number);cout<<endl;return 0;}void convert(int n){int i;char c;if ((i=n/10)!=0)convert(i);c=n%10+'0';cout<<" "<<c;}4.11题#include <iostream>using namespace std;int main(){int f(int);int n,s;cout<<"input the number n:";cin>>n;s=f(n);cout<<"The result is "<<s<<endl;return 0;}int f(int n){;if (n==1)return 1;elsereturn (n*n+f(n-1));}4.12题#include <iostream>#include <cmath>using namespace std;#define S(a,b,c) (a+b+c)/2#define AREA(a,b,c)sqrt(S(a,b,c)*(S(a,b,c)-a)*(S(a,b,c)-b)*(S(a,b,c) -c))int main(){float a,b,c;cout<<"input a,b,c:";cin>>a>>b>>c;if (a+b>c && a+c>b && b+c>a)cout<<"area="<<AREA(a,b,c)<<endl; elsecout<<"It is not a triangle!"<<endl; return 0;}4.14题#include <iostream>using namespace std;//#define LETTER 1int main(){char c;cin>>c;#if LETTERif(c>='a' && c<='z')c=c-32;#elseif(c>='A' && c<='Z')c=c+32;#endifcout<<c<<endl;return 0;}4.15题#include <iostream>using namespace std;#define CHANGE 1int main(){char ch[40];cout<<"input text:"<<endl;;gets(ch);#if (CHANGE){for (int i=0;i<40;i++){if (ch[i]!='\0')if (ch[i]>='a'&& ch[i]<'z'||ch[i]>'A'&& ch[i]<'Z')ch[i]+=1;else if (ch[i]=='z'||ch[i]=='Z')ch[i]-=25;}}#endifcout<<"output:"<<endl<<ch<<endl;return 0;}4.16题file#include <iostream>using namespace std;int a;int main(){extern int power(int);int b=3,c,d,m;cout<<"enter an integer a and its power m:"<<endl;cin>>a>>m;c=a*b;cout<<a<<"*"<<b<<"="<<c<<endl;d=power(m);cout<<a<<"**"<<m<<"="<<d<<endl; return 0;}4.16题fileextern int a;int power(int n){int i,y=1;for(i=1;i<=n;i++)y*=a;return y;}第五章5.1题#include <iostream>#include <iomanip>using namespace std;#include <math.h>int main(){int i,j,n,a[101];for (i=1;i<=100;i++)a[i]=i;a[1]=0;for (i=2;i<sqrt(100);i++)for (j=i+1;j<=100;j++){if(a[i]!=0 && a[j]!=0)if (a[j]%a[i]==0)a[j]=0; }cout<<endl;for (i=1,n=0;i<=100;i++){if (a[i]!=0){cout<<setw(5)<<a[i]<<" ";n++;}if(n==10){cout<<endl;n=0;}}cout<<endl;return 0;}5.2题#include <iostream>using namespace std;//#include <math.h>int main(){int i,j,min,temp,a[11];cout<<"enter data:"<<endl;for (i=1;i<=10;i++){cout<<"a["<<i<<"]=";cin>>a[i]; //输入10个数}cout<<endl<<"The original numbers:"<<endl;;for (i=1;i<=10;i++)cout<<a[i]<<" "; // 输出这10个数cout<<endl;;for (i=1;i<=9;i++) //以下8行是对10个数排序{min=i;for (j=i+1;j<=10;j++)if (a[min]>a[j]) min=j;temp=a[i]; //以下3行将a[i+1]～a[10]中最小者与a[i] 对换a[i]=a[min];a[min]=temp;}cout<<endl<<"The sorted numbers:"<<endl;for (i=1;i<=10;i++) // 输出已排好序的10个数cout<<a[i]<<" ";cout<<endl;return 0;}5.3题#include <iostream>using namespace std;int main(){int a[3][3],sum=0;int i,j;cout<<"enter data:"<<endl;;for (i=0;i<3;i++)for (j=0;j<3;j++)cin>>a[i][j];for (i=0;i<3;i++)sum=sum+a[i][i];cout<<"sum="<<sum<<endl;return 0;}5.4题#include <iostream>using namespace std;int main(){int a[11]={1,4,6,9,13,16,19,28,40,100};int num,i,j;cout<<"array a:"<<endl;for (i=0;i<10;i++)cout<<a[i]<<" ";cout<<endl;;cout<<"insert data:";cin>>num;if (num>a[9])a[10]=num;else。

c程序设计谭浩强习题答案C程序设计是计算机科学与技术专业学生必修的一门课程，谭浩强教授编写的《C程序设计》教材广泛用于教学中，其习题集也是学生学习过程中不可或缺的练习材料。

以下是针对该教材习题的一些参考答案，供参考使用。

第一章：C语言概述1. 问题1：解释C语言的特点。

- 答案：C语言是一种通用的、过程式的编程语言，具有以下特点：简洁、高效、可移植性高、结构化、支持多种编程范式等。

2. 问题2：C语言的发展历史。

- 答案：C语言由丹尼斯·里奇在1972年开发，最初用于UNIX操作系统的编写。

随着UNIX系统的流行，C语言也逐渐被广泛使用并发展。

第二章：数据类型、运算符和表达式1. 问题1：C语言中基本数据类型有哪些？- 答案：C语言中的基本数据类型包括整型（int）、字符型（char）、浮点型（float和double）等。

2. 问题3：解释运算符的优先级。

- 答案：运算符优先级决定了表达式中各运算符的计算顺序。

例如，乘除运算符优先于加减运算符。

第三章：控制语句1. 问题1：解释if语句的用法。

- 答案：if语句用于条件判断，根据条件是否为真来决定执行不同的代码块。

2. 问题2：while循环和for循环的区别是什么？- 答案：while循环在每次迭代前检查条件，而for循环在循环开始前初始化，然后在每次迭代后检查条件。

第四章：数组1. 问题1：一维数组和多维数组的区别。

- 答案：一维数组是线性的，只有一个索引；多维数组可以看作是数组的数组，具有多个索引。

2. 问题2：数组的初始化方法。

- 答案：数组可以通过赋值语句逐个初始化，或者使用初始化列表在声明时初始化。

第五章：指针1. 问题1：指针是什么？- 答案：指针是一种变量，它存储了另一个变量的内存地址。

2. 问题2：指针和数组的关系。

- 答案：数组名可以作为指向数组首元素的指针使用，指针可以用于遍历数组。

第六章：函数1. 问题1：函数的定义和声明。

C程序设计(第四版)(谭浩强)第一章课后习题答案P006 1.1 向屏幕输出文字.#include<stdio.h>//预编译. 代码均调试成功,若有失误大多不是代码问题.自已找找.int main(){printf("Welcome to \n");return 0; //与int main对应,为了程序可移植性,建议全用int main + return 0;.}P008 1.2 求两个数的和.#include<stdio.h>int main(){int a,b,sum;a=5;b=4;sum=a+b;printf("The sum is %d .\n",sum);return 0;}P008 1.3 调用函数比较两个数的大小.#include<stdio.h>int main(){int max(int x,int y); //被调用函数在主函数后面,用前先声明.int a,b,c;scanf("%d,%d",&a,&b); //输入时要按格式来,此处的逗号,用空格会发生错误.c=max(a,b); //a,b作为实参传入被调用函数中.printf("The max is %d .\n",c);return 0;}int max(int x,int y) //定义了两个形参.{int z; //z属于局部变量,可与主函数中相同名字.if (x>y)z=x;elsez=y;return(z); //z作为整个程序的出口值,赋给主函数中的c.}P015 0.6 三个数的大小.(数字0表示课后练习题)#include<stdio.h>int main(){int a,b,c,d; //d是用于存储最大值的.int max(int x , int y , int z); //测试可知,在VS2008中,可以不预先声明.printf("Please input 3 numbers :\n");scanf("%d %d %d",&a,&b,&c);d=max(a,b,c); //调用函数中有三个形参,这里需要传入三个实参,才可运算.printf("The max is :%d .\n",d); // d可以换成max(a,b,c).}int max(int x , int y , int z){int m;if (x>y && x>z) //求三者之大的一种方法.m=x;if (y>x && y>z)m=y;if (z>y && z>x)m=z;return (m); //返回值m给主函数中的d.}C程序设计(第四版)(谭浩强)第2章课后习题答案算法——程序的灵魂P017 2.1 计算机1-5相乘的积.#include<stdio.h>int main(){int i,s=1; //在执行数值操作前一定要先有个初值.for(i=1;i<6;i++) //这里是到6.{s=s*i; //相乘}printf("The sum is %d .\n",s);return 0;}#include<stdio.h> //作出要求:换成1到11间奇数相乘.int main(){int i,s=1; //在执行数值操作前一定要先有个初值.for(i=1;i<12;i++) //这里是到,但题目要求的是取单数.也可以是i=i+2{if(i%2!=0) //i对取模,值为非为奇数;为则为偶数.s=s*i;elsecontinue; //跳过这个for循环的这一次,执行下一次.}printf("The sum is %d .\n",s);return 0;}P019 2.2 按要求输出80分以上的学生信息.暂时没法做.年的概念是地球围绕太阳一周的时间（所谓公转周期）称为一年，这个周期是相当稳定的，很长时间也不会变动1秒，但是真正的一年是365.2423天（目前）。

谭浩强C语言程序设计习题参考答案第一章1.6main(){int a,b,c,max;printf("input three numbers:\n");scanf("%d,%d,%d",&a,&b,&c);max=a;if(max<b)max=b;if(max<c)max=c;printf("max=%d",max);}第二章2.3(1)(10)10=(12)8=(a)16(2)(32)10=(40)8=(20)16(3)(75)10=(113)8=(4b)16(4)(-617)10=(176627)8=(fd97)16(5)(-111)10=(177621)8=(ff91)16(6)(2483)10=(4663)8=(963)16(7)(-28654)10=(110022)8=(9012)16(8)(21003)10=(51013)8=(520b)162.6aabb (8)cc (8)abc(7)AN2.7main(){char c1='C',c2='h',c3='i',c4='n',c5='a';c1+=4, c2+=4, c3+=4, c4+=4, c5+=4;printf("%c%c%c%c%c\n",c1,c2,c3,c4,c5);}2.8main(){int c1,c2;c1=97;c2=98;printf("%c %c",c1,c2);}2.9(1)=2.5(2)=3.52.109,11,9,102.12(1)24 (2)10 (3)60 (4)0 (5)0 (6)0第三章3.4main(){int a,b,c;long int u,n;float x,y,z;char c1,c2;a=3;b=4;c=5;x=1.2;y=2.4;z=-3.6;u=51274;n=128765;c1='a';c2='b';printf("\n");printf("a=%2d b=%2d c=%2d\n",a,b,c);printf("x=%8.6f,y=%8.6f,z=%9.6f\n",x,y,z);printf("x+y=%5.2f y+z=%5.2f z+x=%5.2f\n",x+y,y+z,z+x);printf("u=%6ld n=%9ld\n",u,n);printf("c1='%c'or %d(ASCII)\n",c1,c1);printf("c2='%c'or %d(ASCII)\n",c2,c2);}3.5575 767.856400,-789.12396267.856400,-789.12396267.86 -789.12,67.856400,-789.123962,67.856400,-789.1239626.785640e+001,-7.89e+002A,65,101,411234567,4553207,d68765535,177777,ffff,-1COMPUTER, COM3.6a=3 b=7/x=8.5 y=71.82/c1=A c2=a/3.710 20Aa1.5 -3.75 +1.4,67.8/(空3)10(空3)20Aa1.5(空1)-3.75(空1)(随意输入一个数），67.8回车3.8main(){float pi,h,r,l,s,sq,sv,sz;pi=3.1415926;printf("input r,h\n");scanf("%f,%f",&r,&h);l=2*pi*r;s=r*r*pi;sq=4*pi*r*r;sv=4.0/3.0*pi*r*r*r;sz=pi*r*r*h;printf("l=%6.2f\n",l);printf("s=%6.2f\n",s);printf("sq=%6.2f\n",sq);printf("vq=%6.2f\n",sv);printf("vz=%6.2f\n",sz);}3.9main(){float c,f;scanf("%f",&f);c=(5.0/9.0)*(f-32);printf("c=%5.2f\n",c);}3.10#include"stdio.h"main(){char c1,c2;scanf("%c,%c",&c1,&c2);putchar(c1);putchar(c2);printf("\n");printf("%c%c\n",c1,c2);}第四章4.3(1)0 (2)1 (3)1 (4)0 (5)1 4.4main(){int a,b,c;scanf("%d,%d,%d",&a,&b,&c); if(a<b)if(b<c)printf("max=%d\n",c);elseprintf("max=%d\n",b);else if(a<c)printf("max=%d\n",c);elseprintf("max=%d\n",a);}main(){int a,b,c,temp,max;scanf("%d,%d,%d",&a,&b,&c);temp=(a>b)?a:b;max=(c>temp)?c:temp;printf("max=%d",max);}4.5main(){int x,y;scanf("%d",&x);if(x<1)y=x;else if(x<10)y=2*x-1;else y=3*x-11;printf("y=%d",y);}4.6main(){int score,temp,logic;char grade;logic=1;while(logic){scanf("%d",&score);if(score>=0&&score<=100)logic=0; }if(score==100)temp=9;elsetemp=(score-score%10)/10;switch(temp){case 9:grade='A';break;case 8:grade='B';break;case 7:grade='C';break;case 6:grade='D';break;case 5:case 4:case 3:case 2:case 1:case 0:grade='E';}printf("score=%d,grade=%c",score,grade);}4.7main(){long int num;int indiv,ten,hundred,thousand,ten_thousand,place;scanf("%ld",&num);if(num>9999) place=5;else if(num>999) place=4;else if(num>99) place=3;else if(num>9) place=2;else place=1;printf("place=%d\n",place);ten_thousand=num/10000;thousand=(num-ten_thousand*10000)/1000;hundred=(num-ten_thousand*10000-thousand*1000)/100;ten=(num-ten_thousand*10000-thousand*1000-hundred*100)/10;indiv=num-ten_thousand*10000-thousand*1000-hundred*100-ten*10;switch(place){case 5:printf("%d,%d,%d,%d,%d\n",ten_thousand,thousand,hundred,ten,indiv);printf("%d,%d,%d,%d,%d\n",indiv,ten,hundred,thousand,ten_thousand);break;case 4:printf("%d,%d,%d,%d\n",thousand,hundred,ten,indiv);printf("%d,%d,%d,%d\n",indiv,ten,hundred,thousand);break;case 3:printf("%d,%d,%d\n",hundred,ten,indiv);printf("%d,%d,%d\n",indiv,ten,hundred);break;case 2:printf("%d,%d\n",ten,indiv);printf("%d,%d\n",indiv,ten);break;case 1:printf("%d\n",indiv);printf("%d\n",indiv);}}4.8main(){long i;float bonus,bon1,bon2,bon4,bon6,bon10;bon1=100000*0.1;bon2=bon1+100000*0.075;bon4=bon2+200000*0.05;bon6=bon4+200000*0.03;bon10=bon6+400000*0.015;scanf("%ld",&i);if(i<=1e5)bonus=i*0.1;else if(i<=2e5)bonus=bon1+(i-100000)*0.075; else if(i<=4e5)bonus=bon2+(i-200000)*0.05; else if(i<=6e5)bonus=bon4+(i-400000)*0.03; else if(i<=1e6)bonus=bon6+(i-600000)*0.015; else bonus=bon10+(i-1000000)*0.01;printf("bonus=%10.2f",bonus);}main(){long i;float bonus,bon1,bon2,bon4,bon6,bon10;int branch;bon1=100000*0.1;bon2=bon1+100000*0.075;bon4=bon2+200000*0.05;bon6=bon4+200000*0.03;bon10=bon6+400000*0.015;scanf("%ld",&i);branch=i/100000;if(branch>10)branch=10;switch(branch){case 0:bonus=i*0.1;break;case 1:bonus=bon1+(i-100000)*0.075;break; case 2:case 3:bonus=bon2+(i-200000)*0.05;break; case 4:case 5:bonus=bon4+(i-400000)*0.03;break; case 6:case 7case 8:case 9:bonus=bon6+(i-600000)*0.015;break; case 10:bonus=bon10+(i-1000000)*0.01;}printf("bonus=%10.2f",bonus);}4.9main(){int t,a,b,c,d;scanf("%d,%d,%d,%d",&a,&b,&c,&d);if(a>b){t=a;a=b;b=t;}if(a>c){t=a;a=c;c=t;}if(a>d){t=a;a=d;d=t;}if(b>c){t=b;b=c;c=t;}if(b>d){t=b;b=d;d=t;}if(c>d){t=c;c=d;d=t;}printf("%d %d %d %d\n",a,b,c,d);}4.10main(){int h=10;float x,y,x0=2,y0=2,d1,d2,d3,d4;scanf("%f,%f",&x,&y);d1=(x-x0)*(x-x0)+(y-y0)*(y-y0);d2=(x-x0)*(x-x0)+(y+y0)*(y+y0);d3=(x+x0)*(x+x0)+(y-y0)*(y-y0);d4=(x+x0)*(x+x0)+(y+y0)*(y+y0);if(d1>1&&d2>1&&d3>1&&d4>1)h=0;printf("h=%d",h);}第五章循环控制5.1main(){int a,b,num1,num2,temp;scanf("%d,%d",&num1,&num2);if(num1<num2){temp=num1;num1=num2;num2=temp;} a=num1;b=num2;while(b!=0){temp=a%b;a=b;b=temp;}printf("%d\n",a);printf("%d\n",num1*num2/a);}5.2#include"stdio.h"main(){char c;int letters=0,space=0,digit=0,other=0;while((c=getchar())!='\n'){if(c>='a'&&c<='z'||c>='A'&&c<='Z') letters++;else if(c==' ')space++;else if(c>='0'&&c<='9')digit++;else other++;printf("letters=%d\nspace=%d\ndigit=%d\nother=%d\n",letters,space,digit,other); }5.3main(){int a,n,count=1,sn=0,tn=0;scanf("%d,%d",&a,&n);while(count<=n){tn+=a;sn+=tn;a*=10;++count;}printf("a+aa+aaa+...=%d\n",sn);}5.4main(){float n,s=0,t=1;for(n=1;n<=20;n++){t*=n;s+=t;}printf("s=%e\n",s);}5.5main(){int N1=100,N2=50,N3=10;float k;float s1=0,s2=0,s3=0;for(k=1;k<=N1;k++)s1+=k;for(k=1;k<=N2;k++)s2+=k*k;for(k=1;k<=N3;k++)s3+=1/k;printf("s=%8.2f\n",s1+s2+s3);}5.6main(){int i,j,k,n;for(n=100;n<1000;n++){i=n/100;j=n/10-i*10;k=n%10;if(i*100+j*10+k==i*i*i+j*j*j+k*k*k)printf("n=%d\n",n);}5.7#define M 1000main(){int k0,k1,k2,k3,k4,k5,k6,k7,k8,k9; int i,j,n,s;for(j=2;j<=M;j++){n=0;s=j;for(i=1;i<j;i++){if((j%i)==0){n++;s=s-i;switch(n){case 1:k0=i;break;case 2:k1=i;break;case 3:k2=i;break;case 4:k3=i;break;case 5:k4=i;break;case 6:k5=i;break;case 7:k6=i;break;case 8:k7=i;break;case 9:k8=i;break;case 10:k9=i;break;}}}if(s==0){printf("j=%d\n",j);if(n>1)printf("%d,%d",k0,k1);if(n>2)printf(",%d",k2);if(n>3)printf(",%d",k3);if(n>4)printf(",%d",k4);if(n>5)printf(",%d",k5);if(n>6)printf(",%d",k6);if(n>7)printf(",%d",k7);if(n>8)printf(",%d",k8);if(n>9)printf(",%d\n",k9);}}}main(){static int k[10];int i,j,n,s;for(j=2;j<=1000;j++) {n=-1;s=j;for(i=1;i<j;i++){if((j%i)==0){n++;s=s-i;k[n]=i;}}if(s==0){printf("j=%d\n",j);for(i=0;i<n;i++)printf("%d,",k[i]);printf("%d\n",k[n]);}}}5.8main(){int n,t,number=20;float a=2;b=1;s=0;for(n=1;n<=number;n++) {s=s+a/b;t=a,a=a+b,b=t;}printf("s=%9.6f\n",s);}5.9main(){float sn=100.0,hn=sn/2; int n;for(n=2;n<=10;n++) {sn=sn+2*hn;hn=hn/2;}printf("sn=%f\n",sn);printf("hn=%f\n",hn);}5.10main(){int day,x1,x2;day=9;x2=1;while(day>0){x1=(x2+1)*2;x2=x1;day--;}printf("x1=%d\n",x1);}5.11#include"math.h"main(){float a,xn0,xn1;scanf("%f",&a);xn0=a/2;xn1=(xn0+a/xn0)/2;do{xn0=xn1;xn1=(xn0+a/xn0)/2;}while(fabs(xn0-xn1)>=1e-5);printf("a=%5.2f\n,xn1=%8.2f\n",a,xn1); }5.12#include"math.h"main(){float x,x0,f,f1;x=1.5;do{x0=x;f=((2*x0-4)*x0+3)*x0-6;f1=(6*x0-8)*x0+3;x=x0-f/f1;}while(fabs(x-x0)>=1e-5);printf("x=%6.2f\n",x);}5.13#include"math.h"main(){float x0,x1,x2,fx0,fx1,fx2;do{scanf("%f,%f",&x1,&x2);fx1=x1*((2*x1-4)*x1+3)-6;fx2=x2*((2*x2-4)*x2+3)-6;}while(fx1*fx2>0);do{x0=(x1+x2)/2;fx0=x0*((2*x0-4)*x0+3)-6;if((fx0*fx1)<0){x2=x0;fx2=fx0;}else{x1=x0;fx1=fx0;}}while(fabs(fx0)>=1e-5);printf("x0=%6.2f\n",x0);}5.14main(){int i,j,k;for(i=0;i<=3;i++){for(j=0;j<=2-i;j++)printf(" ");for(k=0;k<=2*i;k++)printf("*");printf("\n");}for(i=0;i<=2;i++){for(j=0;j<=i;j++)printf(" ");for(k=0;k<=4-2*i;k++)printf("*");printf("\n");}}5.15main(){char i,j,k;for(i='x';i<='z';i++)for(j='x';j<='z';j++){if(i!=j)for(k='x';k<='z';k++){if(i!=k&&j!=k){if(i!='x'&&k!='x'&&k!='z')printf("\na--%c\tb--%c\tc--%c\n",i,j,k);}}}}第六章数组6.1#include <math.h>#define N 101main(){ int i,j,line,a[N];for (i=2;i<N;i++) a[i]=i; for (i=2;i<sqrt(N);i++)for (j=i+1;j<N;j++){if(a[i]!=0 && a[j]!=0)if (a[j]%a[i]==0)a[j]=0; }printf("\n");for (i=2,line=0;i<N;i++) { if(a[i]!=0){ printf("%5d",a[i]);line++; }if(line==10){ printf("\n");line=0; }}}6.2#define N 10main(){int i,j,min,temp,a[N];for(i=0;i<N;i++)scanf("%d",&a[i]);for(i=0;i<N-1;i++){min=i;for(j=i+1;j<N;j++)if(a[min]>a[j])min=j;temp=a[i];a[i]=a[min];a[min]=temp;}for(i=0;i<N;i++)printf("%5d",a[i]);}6.3main(){float a[3][3],sum;int i,j;for(i=0;i<3;i++)for(j=0;j<3;j++){scanf("%f",&sum);a[i][j]=sum;}for(i=0;i<3;i++)sum=sum+a[i][i];printf("sum=%f",sum);}6.4main(){int a[11]={1,4,6,9,13,16,19,28,40,100}; int temp1,temp2,number, end,i,j;scanf("%d",&number);end=a[9];if(number>end) a[10]=number;else{for(i=0;i<10;i++){if(a[i]>number){temp1=a[i];a[i]=number;for(j=i+1;j<11;j++){temp2=a[j];a[j]=temp1;temp1=temp2;}break;}}}for(i=0;i<11;i++)printf("%6d",a[i]);}6.5#define N 5main(){int a[N]={8,6,5,4,1},i,temp;for(i=0;i<N/2;i++){temp=a[i];a[i]=a[N-i-1];a[N-i-1]=temp;}for(i=0;i<N;i++)printf("%4d",a[i]);}6.6#define N 11main(){int i,j,a[N][N];for(i=1;i<N;i++){a[i][i]=1;a[i][1]=1;}for(i=3;i<N;i++)for(j=2;j<i;j++)a[i][j]=a[i-1][j-1]+a[i-1][j];for(i=1;i<N;i++){for(j=1;j<=i;j++)printf("%6d",a[i][j]);printf("\n");}}6.7main(){int a[16][16],i,j,k,p,m,n;p=1;while(p==1){scanf("%d",&n);if((n!=0)&&(n<=15)&&(n%2!=0))p=0; }for(i=1;i<=n;i++)for(j=1;j<=n;j++)a[i][j]=0;j=n/2+1;a[1][j]=1;for(k=2;k<=n*n;k++){i=i-1;j=j+1;if((i<1)&&(j>n)){i=i+2;j=j-1;}else{if(i<1)i=n;if(j>n)j=1;}if(a[i][j]==0)a[i][j]=k;else{i=i+2;j=j-1;a[i][j]=k;}}for(i=1;i<=n;i++){for(j=1;j<=n;j++)printf("%3d",a[i][j]);printf("\n");}}6.8#define N 10#define M 10main(){int i,j,k,m,n,flag1,flag2,a[N][M],max,maxi,maxj; scanf("%d,%d",&n,&m);for(i=0;i<n;i++)for(j=0;j<m;j++)scanf("%d",&a[i][j]);flag2=0;for(i=0;i<n;i++){max=a[i][0];for(j=0;j<m;j++)if(max<a[i][j]){max=a[i][j];maxj=j;}for(k=0,flag1=1;k<n&&flag1;k++)if(max>a[k][maxj])flag1=0;if(flag1){ printf("\na[%d][%d]=%d\n",i,maxj,max);flag2=1;}}if(!flag2) printf("NOT");}6.9#include<stdio.h>#define N 15main(){int i,j,number,top,bott,min,loca,a[N],flag;char c;for(i=0;i<=N;i++)scanf("%d",&a[i]);flag=1;while(flag){scanf("%d",&number);loca=0;top=0;bott=N-1;if((number<a[0])||(number>a[N-1]))loca=-1;while((loca==0)&&(top<=bott)){min=(bott+top)/2;if(number==a[min]){loca=min;printf("number=%d,loca=%d\n",number,loca+1);}else if(number<a[min])bott=min-1;elsetop=min+1;}if(loca==0||loca==-1)printf("%d not in table\n",number);printf("continue Y/N or y/n\n");c=getchar();if(c=='N'||c=='n')flag=0;}}6.10main(){int i,j,uppn,lown,dign,span,othn;char text[3][80];uppn=lown=dign=span=othn=0;for(i=0;i<3;i++){gets(text[i]);for(j=0;j<80&&text[i][j]!='\0';j++){if(text[i][j]>='A'&&text[i][j]<='Z')uppn++;else if(text[i][j]>='a'&&text[i][j]<='z')lown++;else if(text[i][j]>='0'&&text[i][j]<='9')dign++;else if(text[i][j]==' ')span++;elseothn++;}}for(i=0;i<3;i++)printf("%s\n",text[i]);printf("uppn=%d\n",uppn);printf("lown=%d\n",lown);printf("dign=%d\n",dign);printf("span=%d\n",span);printf("othn=%d\n",othn);}6.11main(){static char a[5]={'*','*','*','*','*'};int i,j,k;char space=' ';for(i=0;i<=5;i++){printf("\n");for(j=1;j<=3*i;j++)printf("%1c",space);for(k=0;k<=5;k++)printf("%3c",a[k]);}}6.12#include<stdio.h>main(){int i,n;char ch[80],tran[80];gets(ch);i=0;while(ch[i]!='\0'){if((ch[i]>='A')&&(ch[i]<='Z'))tran[i]=26+64-ch[i]+1+64;else if((ch[i]>='a')&&(ch[i]<='z'))tran[i]=26+96-ch[i]+1+96;elsetran[i]=ch[i];i++;}n=i;for(i=0;i<n;i++)putchar(tran[i]);}6.13main(){char s1[80],s2[40];int i=0,j=0;scanf("%s",s1);scanf("%s",s2);while(s1[i]!='\0')i++;while(s2[j]!='\0')s1[i++]=s2[j++];s1[i]='\0';printf("s=%s\n",s1);}6.14#include<stdio.h>main(){int i,resu;char s1[100],s2[100];gets(s1);gets(s2);i=0;while((s1[i]==s2[i])&&(s1[i]!='\0'))i++;if(s1[i]=='\0'&&s2[i]=='\0')resu=0;elseresu=s1[i]-s2[i];printf("s1=%s,s2=%s,resu=%d\n",s1,s2,resu); }6.15#include"stdio.h"main(){char from[80],to[80];;int i;scanf("%s",from);for(i=0;i<=strlen(from);i++)to[i]=from[i];printf("%s\n",to);}第七章7.1hcf(u,v)int u,v;{int a,b,t,r;if(u>v){t=u;u=v;v=t;}a=u;b=v;while((r=b%a)!=0){b=a;a=r;}return(a);}lcd(u,v,h)int u,v,h;{return(u*v/h);}main(){int u,v,h,l;scanf("%d,%d",&u,&v);h=hcf(u,v);printf("H.C.F=%d\n",h);l=lcd(u,v,h);printf("L.C.D=%d\n",l);}7.2#include"math.h"float x1,x2,disc,p,q;greater_than_zero(a,b)float a,b;{x1=(-b+sqrt(disc))/(2*a);x2=(-b-sqrt(disc))/(2*a);}equal_to_zero(a,b)flaot a,b;{x1=x2=-b/(2*a);}smaller_than_zero(a,b)float a,b;{p=-b/(2*a);q=sqrt(-disc)/(2*a);}main(){float a,b,c;scanf("%f,%f,%f",&a,&b,&c);disc=b*b-4*a*c;if(fabs(disc)<=1e-5){equal_to_zero(a,b);printf("x1=%5.2f\tx2=%5.2f\n",x1,x2); }else if(disc>0){greater_than_zero(a,b);printf("x1=%5.2f\tx2=%5.2f\n",x1,x2);}else{smaller_than_zero(a,b);printf("x1=%5.2f+%5.2fi\tx2=%5.2f-%5.2fi\n",p,q,p,q); }}7.3main(){int number;scanf("%d",&number);if(prime(number))printf("yes");elseprintf("no");}int prime(number)int number;{int flag=1,n;for(n=2;n<number/2&&flag==1;n++)if(number%n==0)flag=0;return(flag);}7.4#define N 3int array[N][N];convert(array)int array[3][3];{int i,j,t;for(i=0;i<N;i++)for(j=i+1;j<N;j++){t=array[i][j];array[i][j]=array[j][i];array[j][i]=t;}}main(){int i,j;for(i=0;i<N;i++)for(j=0;j<N;j++)scanf("%d",&array[i][j]);convert(array);for(i=0;i<N;i++){printf("\n");for(j=0;j<N;j++)printf("%5d",array[i][j]);}}7.5main(){char str[100];scanf("%s",str);inverse(str);printf("%s\n",str);}inverse(str)char str[];{char t;int i,j;for(i=0,j=strlen(str);i<strlen(str)/2;i++,j--) {t=str[i];str[i]=str[j-1];str[j-1]=t;}}7.6char concate(str1,str2,str)char str1[],str2[],str[];{int i,j;for(i=0;str1[i]!='\0';i++)str[i]=str1[i];for(j=0;str2[j]!='\0';j++)str[i+j]=str2[j];str[i+j]='\0';}main(){char s1[100],s2[100],s[100];scanf("%s",s1);scanf("%s",s2);concate(s1,s2,s);printf("\ns=%s",s);}7.7main(){char str[80],c[80];void cpy();gets(str);cpy(str,c);printf("\n%s\n",c);}void cpy(s,c)char s[],c[];{int i,j;for(i=0,j=0;s[i]!='\0';i++)if(s[i]=='a'||s[i]=='A'||s[i]=='e'||s[i]=='E'||s[i]=='i'||s[i]=='I'||s[i]=='o'||s[i]=='O'||s[i]=='u'||s[i]=='U'){c[j]=s[i];j++;}c[j]='\0';}7.8main(){char str[80];scanf("%s",str);insert(str);}insert(str)char str[];{int i;for(i=strlen(str);i>0;i--){str[i*2]=str[i];str[i*2-1]=' ';}printf("%s\n",str);}7.9int alph,digit,space,others;main(){char text[80];gets(text);alph=0,digit=0,space=0,others=0;count(text);printf("\nalph=%d,digit=%d,space=%d,others=%d\n",alph,digit,space,others); }count(str)char str[];{int i;for(i=0;str[i]!='\0';i++)if((str[i]>='a'&&str[i]<='z')||(str[i]>='A'&&str[i]<='Z'))alph++;else if(str[i]>='0'&&str[i]<='9')digit++;else if(strcmp(str[i],' ')==0)space++;elseothers++;}7.10int alph(c)char c;{if((c>='a'&&c<='z')||(c>='A'&&c<='Z')) return(1);elsereturn(0);}int longest(string)char string[];{int len=0,i,length=0,flag=1,place,point; for(i=0;i<=strlen(string);i++)if(alph(string[i]))if(flag){point=i;flag=0;}elselen++;else{flag=1;if(len>length){length=len;place=point;len=0;}}return(place);}main(){int i;char line[100];gets(line);for(i=longest(line);alph(line[i]);i++) printf("%c",line[i]);printf("\n");}7.11#define N 10char str[N];{int i,flag;for(flag=1;flag==1;){scanf("%s",str);if(strlen(str)>N)printf("input error");elseflag=0;}sort(str);for(i=0;i<N;i++)printf("%c",str[i]);}sort(str)char str[N];{int i,j;char t;for(j=1;j<N;j++)for(i=0;(i<N-j)&&(str[i]!='\0');i++)if(str[i]>str[i+1]){t=str[i];str[i]=str[i+1];str[i+1]=t;}}7.12#include<math.h>float solut(a,b,c,d)float a,b,c,d;{float x=1,x0,f,f1;do{x0=x;f=((a*x0+b)*x0+c)*x0+d;f1=(3*a*x0+2*b)*x0+c;x=x0-f/f1;}while(fabs(x-x0)>=1e-5);return(x);}main(){float a,b,c,d;scanf("%f,%f,%f,%f",&a,&b,&c,&d); printf("x=%10.7f\n",solut(a,b,c,d)); }main(){int x,n;float p();scanf("%d,%d",&n,&x);printf("P%d(%d)=%10.2f\n",n,x,p(n,x));}float p(tn,tx)int tn,tx;{if(tn==0)return(1);else if(tn==1)return(tx);elsereturn(((2*tn-1)*tx*p((tn-1),tx)-(tn-1)*p((tn-2),tx))/tn); }7.14#define N 10#define M 5float score[N][M];float a_stu[N],a_cor[M];main(){int i,j,r,c;float h;float s_diff();float highest();r=0;c=1;input_stu();avr_stu();avr_cor();printf("\n number class 1 2 3 4 5 avr");for(i=0;i<N;i++){printf("\nNO%2d",i+1);for(j=0;j<M;j++)printf("%8.2f",score[i][j]);printf("%8.2f",a_stu[i]);}printf("\nclassavr");for(j=0;j<M;j++)printf("%8.2f",a_cor[j]);h=highest(&r,&c);printf("\n\n%8.2f %d %d\n",h,r,c);printf("\n %8.2f\n",s_diff());input_stu(){int i,j;float x;for(i=0;i<N;i++){for(j=0;j<M;j++){scanf("%f",&x);score[i][j]=x;}}}avr_stu(){int i,j;float s;for(i=0;i<N;i++){for(j=0,s=0;j<M;j++)s+=score[i][j];a_stu[i]=s/5.0;}}avr_cor(){int i,j;float s;for(j=0;j<M;j++){for(i=0,s=0;i<N;i++)s+=score[i][j];a_cor[j]=s/(float)N; }}float highest(r,c)int *r,*c;{float high;int i,j;high=score[0][0];for(i=0;i<N;i++)for(j=0;j<M;j++)if(score[i][j]>high){high=score[i][j];*r=i+1;*c=j+1;}return(high);}float s_diff()float sumx=0.0,sumxn=0.0;for(i=0;i<N;i++){sumx+=a_stu[i]*a_stu[i];sumxn+=a_stu[i];}return(sumx/N-(sumxn/N)*(sumxn/N)); }7.15#include<stdio.h>#define N 10void input_e(num,name)int num[];char name[N][8];{int i;for(i=0;i<N;i++){scanf("%d",&num[i]);gets(name[i]);}}void sort(num,name)int num[];char name[N][8];{int i,j,min,temp1;char temp2[8];for(i=0;i<N-1;i++){min=i;for(j=i;j<N;j++)if(num[min]>num[j])min=j;temp1=num[i];num[i]=num[min];num[min]=temp1;strcpy(temp2,name[i]);strcpy(name[i],name[min]);strcpy(name[min],temp2);}for(i=0;i<N;i++)printf("\n%5d%10s",num[i],name[i]); }void search(n,num,name)int n,num[];char name[N][8];{int top,bott,min,loca;loca=0;bott=N-1;if((n<num[0])||(n>num[N-1]))loca=-1;while((loca==0)&&(top<=bott)){min=(bott+top)/2;if(n==num[min]){loca=min;printf("number=%d,name=%s\n",n,name[loca]);}else if(n<num[min])bott=min-1;elsetop=min+1;}if(loca==0||loca==-1)printf("number=%d is not in table\n",n);}main(){int num[N],number,flag,c,n;char name[N][8];input_e(num,name);sort(num,name);for(flag=1;flag;){scanf("%d",&number);search(number,num,name);printf("continue?Y/N!");c=getchar();if(c=='N'||c=='n')flag=0;}}7.16#include<stdio.h>#define MAX 1000main(){int c,i,flag,flag1;char t[MAX];i=0;flag=0;flag1=1;while((c=getchar())!='\0'&&i<MAX&&flag1){if(c>='0'&&c<='9'||c>='A'&&c<='F'||c>='a'&&c<='f') {flag=1;t[i++]=c;}else if(flag){t[i]='\0';printf("\nnumber=%d\n",htoi(t));printf("continue?");c=getchar();if(c=='n'||c=='N')flag1=0;else{flag=0;i=0;}}}}htoi(s)char s[];{int i,n;n=0;for(i=0;s[i]!='\0';i++){if(s[i]>='0'&&s[i]<='9')n=n*16+s[i]-'0';if(s[i]>='a'&&s[i]<='f')n=n*16+s[i]-'a'+10;if(s[i]>='A'&&s[i]<='F')n=n*16+s[i]-'A'+10;}return(n);}7.17#include<stdio.h>void convert(n)int n;{int i;if((i=n/10)!=0)convert(i);putchar(n%10+'0');}main(){int number;scanf("%d",&number);if(number<0){putchar('-');number=-number;}convert(number);}7.18main(){int year,month,day;int days;scanf("\n%d,%d,%d",&year,&month,&day);days=sum_day(month,day);if(leap(year)&&(month>=3))days+=1;printf("days=%d\n",days);}static int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31} int sum_day(month,day)int month,day;{int i;for(i=1;i<month;i++)day+=day_tab[i];return(day);}int leap(year)int year;{int leap;leap=year%4==0&&year%100!=0||year%400==0;return(leap);}第八章编译预处理8.1#define SW AP(a,b) t=b;b=a;a=tmain(){int a,b,t;scanf("%d,%d",&a,&b);SWAP(a,b);printf("a=%d\tb=%d\n",a,b);}8.2#define SURPLUS(a,b) ((a)%(b))main(){int a,b;scanf("%d,%d",&a,&b);printf("%d",SURPLUS(a,b));}。

5.2#include<stdio.h>#include<math.h>int main(){int sign=1,i=0;double pi=0.0,n=1.0,term=1.0; while(fabs(term)>=1e-6){pi=pi+term;n=n+2;sign=-sign;term=sign/n;i++;}pi=pi*4;printf("i=%d\npi=%10.8f\n",i,pi); return 0;}5.2#include<stdio.h>#include<math.h>int main(){int sign=1,i=0;double pi=0.0,n=1.0,term=1.0; while(fabs(term)>=1e-8){pi=pi+term;n=n+2;sign=-sign;term=sign/n;i++;}pi=pi*4;printf("i=%d\npi=%10.8f\n",i,pi); return 0;}5.3#include<stdio.h>int main(){int m,n,t,q,r;printf("输入两个正整数："); scanf("%d,%d",&m,&n);if(m<n){ t=m;m=n;n=t;}q=m*n;while(n!=0){ r=m%n;m=n;n=r;}printf("最大公约数：%d\n",m); printf("最小公倍数：%d\n",q/m); yreturn 0;}5.4#include<stdio.h>int main(){int letter=0,space=0,digit=0,other=0;char c;printf("输入一行字符：\n");while((c=getchar())!='\n'){if(c>='a'&&c<='z'||c>='A'&&c<='Z')letter++;else if(c==' ')space++;else if(c>='0'&&c<='9')digit++;elseother++;}printf("英文字母个数：%d\n空格个数：%d\n 数字个数：%d\n其他：%d\n",letter,space,digit,other);return 0;}5.5#include<stdio.h>int main(){int n,i,a,t=0,s=0;printf("input a,n=");scanf("%d,%d",&a,&n);for(i=1;i<=n;i++){t=t+a;s=s+t;a=10*a;}printf("a+aa+aaa+……+aa……a=%d\n",s); return 0;}5.6#include<stdio.h>int main(){double n,t=1,s=0;for(n=1;n<=20;n++){t=t*n;s=s+t;}printf("结果为：%25.15e",s); return 0;}5.7#include<stdio.h>int main(){float s3=0,k;int s1=0,s2=0;for(k=1;k<=100;k++){s1=s1+k;}for(k=1;k<=50;k++){s2=s2+k*k;}for(k=1;k<=10;k++){s3=1/k+s3;}printf("%d\n%d\n%15.6f\n%15.6f\n",s1,s2,s3, s1+s2+s3);return 0;}5.8#include<stdio.h>int main(){int a,b,s,g;for(a=100;a<=999;a++){b=a/100;s=(a-b*100)/10;g=a-b*100-s*10;if(a==b*b*b+s*s*s+g*g*g)printf("%d\n",a);}printf("\n");return 0;}5.9#include<stdio.h>int main(){ int m,s,i;for(m=2;m<1000;m++){ s=0;for(i=1;i<m;i++)if((m%i)==0)s=s+i;if(s==m){ printf("%d its factors are",m);for(i=1;i<m;i++)if(m%i==0) printf("%d ",i);printf("\n");}}return 0;}5.10#include<stdio.h>int main(){float a=2,b=1,t,s=0;int i;for(i=1;i<=20;i++){ s=s+a/b;t=a;a=a+b;b=t;}printf("sum=%15.12f\n",s);return 0;}5.11#include<stdio.h>int main(){int i;float s=0,l=100,h=l/2;for(i=1;i<=10;i++){s=s+2*h;h=h/2;}printf("共经过：%f\n第10次反弹：%f\n",s,h);return 0;}12.#include<stdio.h>int main(){int i,a=1;for(i=9;i>0;i--)a=2*a+2;printf("%d\n",a);return 0;}#include<stdio.h>int main(){int i,j,k;for(i=0;i<=3;i++){ for(j=0;j<3-i;j++)printf(" ");for(k=0;k<=2*i;k++)printf("*");printf("\n");}for(i=0;i<=2;i++){ for(j=0;j<=i;j++)printf(" ");for(k=0;k<=4-2*i;k++)printf("*");printf("\n");}return 0;}#include <stdio.h>int main(){char i,j,k;for(i='x';i<='z';i++)for(j='x';j<='z';j++)if(i!=j)for(k='x';k<='z';k++)if(i!=k&&j!=k)if(i!='x'&&k!='x'&&k!='z')printf("A-%c\nB-%c\nC-%c\n"<i,j,k); return 0;}。

第一章程序设计和C语言【第15页】1-5#include <stdio.h>int main ( ){ printf ("**************************\n\n");printf(" Very Good!\n\n");printf ("**************************\n");return 0;}1-6#include <stdio.h>int main(){int a,b,c,max;printf("please input a,b,c:\n");scanf("%d,%d,%d",&a,&b,&c);max=a;if (max<b)max=b;if (max<c)max=c;printf("The largest number is %d\n",max);return 0;}第2章算法——程序的灵魂【第36页】暂无答案第3章最简单的C程序设计——顺序程序设计【第82页】3-1#include <stdio.h>#include <math.h>int main(){float p,r,n;r=0.1;n=10;p=pow(1+r,n);printf("p=%f\n",p);return 0;}3-2-1#include <stdio.h>#include <math.h>int main(){float r5,r3,r2,r1,r0,p,p1,p2,p3,p4,p5;p=1000;r1=0.0414;r0=0.0072;p1=p*((1+r5)*5); // 一次存5年期p2=p*(1+2*r2)*(1+3*r3); // 先存2年期，到期后将本息再存3年期p3=p*(1+3*r3)*(1+2*r2); // 先存3年期，到期后将本息再存2年期p4=p*pow(1+r1,5); // 存1年期，到期后将本息存再存1年期，连续存5次p5=p*pow(1+r0/4,4*5); // 存活期存款。

第五章循环控制5.3输入两个正整数m和n，求其最大公约数和最小公倍数。

main(){long m,n,i=1,j,s;scanf("%ld,%ld",&m,&n);for(;i<=m&&i<=n;i++){if(m%i==0&&n%i==0) s=i;}if(m>=n) j=m;else j=n;for(;!(j%m==0&&j%n==0);j++);printf("s=%ld,j=%ld\n",s,j);}5.4输入一行字符，分别统计出其中英文字母、空格、数字和其他字符的个数。

#include"stdio.h"main(){char c;int i=0,j=0,k=0,l=0;while((c=getchar())!=’\n’){if(c>=65&&c<=90||c>=97&&c<=122) i++;else if(c>=48&&c<=57) j++;else if(c==32) k++;else l++;}printf("i=%d,j=%d,k=%d,l=%d\n",i,j,k,l);}5.5求Sn=a+aa+aaa+…+aa…aaa（有n个a）之值，其中a是一个数字。

例如：2+22+222+2222+22222（n=5），n由键盘输入。

#include"math.h"main(){int n,sum=0,i=1,s=2;scanf("%d",&n);while(i<=n){sum=sum+s;s=s+2*pow(10,i);i++;}printf("sum=%d\n",sum);}5.6求，（即求1!+2!+3!+4!+5!+…+20!）main(){int n,i=1;long sum=0,s=1;scanf("%d",&n);while(i<=n) {s=s*i;sum=sum+s;i++;}printf("sum=%ld\n",sum);}5.7求main(){double i=1,j=1,k=1,s1=0,s2=0,s3=0,sum;for(;i<=100;i++) s1=s1+i;for(;j<=50;j++) s2=s2+j*j;for(;k<=10;k++) s3=s3+1/k;sum=s1+s2+s3;printf("sum=%f\n",sum);}5.8打印出所有"水仙花数"，所谓"水仙花数"是指一个三位数，其各位数字立方和等于该本身。

C程序设计(第四版)（谭浩强）第一章课后习题答案P006 1。

1 向屏幕输出文字。

＃include<stdio。

h>//预编译. 代码均调试成功,若有失误大多不是代码问题.自已找找.int main（)｛printf（"Welcome to mybbing。

com\n"）;return 0；//与int main对应,为了程序可移植性，建议全用int main + return 0；。

}P008 1。

2 求两个数的和.＃include〈stdio.h〉int main()｛int a，b，sum；a=5；b=4；sum=a+b；printf（"The sum is ％d .\n",sum)；return 0；}P008 1。

3 调用函数比较两个数的大小。

＃include<stdio.h〉int main()｛int max(int x，int y）; //被调用函数在主函数后面,用前先声明.int a，b，c;scanf（”%d，%d",＆a，＆b）；//输入时要按格式来，此处的逗号,用空格会发生错误.c=max(a,b）; //a,b作为实参传入被调用函数中.printf（”The max is %d 。

\n",c）；return 0;}int max（int x，int y) //定义了两个形参。

｛int z; //z属于局部变量,可与主函数中相同名字。

if (x>y）z=x；elsez=y;return（z）；//z作为整个程序的出口值,赋给主函数中的c.}P015 0。

6 三个数的大小.(数字0表示课后练习题)#include〈stdio。

h>int main（）{int a,b,c,d；//d是用于存储最大值的。

int max(int x ，int y ，int z）；//测试可知，在VS2008中,可以不预先声明.printf(”Please input 3 numbers ：\n"）;scanf（”％d %d %d",&a,＆b,&c）;d=max（a，b,c）；//调用函数中有三个形参，这里需要传入三个实参，才可运算.printf("The max is ：％d .\n”，d); // d可以换成max(a,b，c)。

1-5 #include <>int main ( ){ printf ("**************************\n\n"); printf(" Very Good!\n\n");printf ("**************************\n"); return 0;}1-6#include <>int main(){int a,b,c,max;printf("please input a,b,c:\n");scanf("%d,%d,%d",&a,&b,&c);max=a;if (max<b)max=b;if (max<c)max=c;printf("The largest number is %d\n",max);return 0;}3-1 #include <>#include <>int main(){float p,r,n;r=;n=10;p=pow(1+r,n);printf("p=%f\n",p);return 0;}3-2-1#include <>#include <>int main(){float r5,r3,r2,r1,r0,p,p1,p2,p3,p4,p5;p=1000;r5=;r3=;r2=;r1=;r0=;p1=p*((1+r5)*5); #include <>#include <>int main(){float d=300000,p=6000,r=,m;m=log10(p/(p-d*r))/log10(1+r);printf("m=%\n",m);return 0;}3-4#include <>int main(){int c1,c2;c1=197;c2=198;printf("c1=%c,c2=%c\n",c1,c2);printf("c1=%d，c2=%d\n",c1,c2);return 0;}3-5#include <>int main(){int a,b;float x,y;char c1,c2;scanf("a=%d b=%d",&a,&b);scanf("%f %e",&x,&y);scanf("%c%c",&c1,&c2);printf("a=%d,b=%d,x=%f,y=%f,c1=%c,c2=%c\n",a,b,x,y,c1,c2); return 0;}3-6#include <>int main(){char c1='C',c2='h',c3='i',c4='n',c5='a';c1=c1+4;c2=c2+4;c3=c3+4;c4=c4+4;c5=c5+4;printf("passwor is %c%c%c%c%c\n",c1,c2,c3,c4,c5);return 0;}3-7#include <>int main (){float h,r,l,s,sq,vq,vz;float pi=;printf("请输入圆半径r，圆柱高h∶");scanf("%f,%f",&r,&h); #include <> int main(){ int x,y;printf("输入x:");scanf("%d",&x);if(x<1) /* x<1 */{ y=x;printf("x=%3d, y=x=%d\n" ,x,y);}else if(x<10) /* 1=<x<10 */{ y=2*x-1;printf("x=%d, y=2*x-1=%d\n",x,y);}else /* x>=10 */{ y=3*x-11;printf("x=%d, y=3*x-11=%d\n",x,y);}return 0;}4-8#include <>int main(){ float score;char grade;printf("请输入学生成绩:");scanf("%f",&score);while (score>100||score<0){printf("\n 输入有误,请重输");scanf("%f",&score);}switch((int)(score/10)){case 10:case 9: grade='A';break;case 8: grade='B';break;case 7: grade='C';break;case 6: grade='D';break;case 5:case 4:case 3:case 2:case 1:case 0: grade='E';}printf("成绩是 %,相应的等级是%c\n ",score,grade);return 0;}4-10-1#include <>int main(){int i;double bonus,bon1,bon2,bon4,bon6,bon10;bon1=100000*;bon2=bon1+100000*;bon4=bon2+100000*;bon6=bon4+100000*;bon10=bon6+400000*;printf("请输入利润i:");scanf("%d",&i);if (i<=100000)bonus=i*;else if (i<=200000)bonus=bon1+(i-100000)*;else if (i<=400000)bonus=bon2+(i-200000)*;else if (i<=600000)bonus=bon4+(i-400000)*;else if (i<=1000000)bonus=bon6+(i-600000)*;elsebonus=bon10+(i-1000000)*;printf("奖金是: %\n",bonus);return 0;}4-10-2#include <>int main(){int i;double bonus,bon1,bon2,bon4,bon6,bon10;int branch;bon1=100000*;bon2=bon1+100000*;bon4=bon2+200000*;bon6=bon4+200000*;bon10=bon6+400000*;printf("请输入利润i:");scanf("%d",&i);branch=i/100000;if (branch>10) branch=10;switch(branch){ case 0:bonus=i*;break;case 1:bonus=bon1+(i-100000)*;break; case 2:case 3: bonus=bon2+(i-200000)*;break; case 4:case 5: bonus=bon4+(i-400000)*;break; case 6:case 7:case 8:case 9: bonus=bon6+(i-600000)*;break; case 10: bonus=bon10+(i-1000000)*;}printf("奖金是 %\n",bonus);return 0;}4-11#include <>int main(){int t,a,b,c,d;printf("请输入四个数:");scanf("%d,%d,%d,%d",&a,&b,&c,&d);printf("a=%d,b=%d,c=%d,d=%d\n",a,b,c,d); if (a>b){ t=a;a=b;b=t;}if (a>c){ t=a;a=c;c=t;}if (a>d){ t=a;a=d;d=t;}if (b>c){ t=b;b=c;c=t;}if (b>d){ t=b;b=d;d=t;}if (c>d){ t=c;c=d;d=t;}printf("排序结果如下: \n");printf("%d %d %d %d \n" ,a,b,c,d); return 0;}4-12#include <>int main(){int h=10;float x1=2,y1=2,x2=-2,y2=2,x3=-2,y3=-2,x4=2,y4=-2,x,y,d1,d2,d3,d4; printf("请输入一个点(x,y):");scanf("%f,%f",&x,&y);d1=(x-x4)*(x-x4)+(y-y4)*(y-y4); /*求该点到各中心点距离*/ d2=(x-x1)*(x-x1)+(y-y1)*(y-y1);d3=(x-x2)*(x-x2)+(y-y2)*(y-y2);d4=(x-x3)*(x-x3)+(y-y3)*(y-y3);if (d1>1 && d2>1 && d3>1 && d4>1) h=0; /*判断该点是否在塔外*/ printf("该点高度为 %d\n",h);return 0;}5-3#include <>int main(){int p,r,n,m,temp;printf("请输入两个正整数n,m:");scanf("%d,%d,",&n,&m);if (n<m){temp=n;n=m;m=temp;}p=n*m;while(m!=0){r=n%m;n=m;m=r;}printf("它们的最大公约数为:%d\n",n);printf("它们的最小公约数为:%d\n",p/n);return 0;}5-4#include <>int main(){char c;int letters=0,space=0,digit=0,other=0;printf("请输入一行字符:\n");while((c=getchar())!='\n'){if (c>='a' && c<='z' || c>='A' && c<='Z')letters++;else if (c==' ')space++;else if (c>='0' && c<='9')digit++;elseother++;}printf("字母数:%d\n空格数:%d\n数字数:%d\n其它字符数:%d\n",letters,space,digit,other);return 0;}5-5#include <>int main(){int a,n,i=1,sn=0,tn=0;printf("a,n=:");scanf("%d,%d",&a,&n);while (i<=n){tn=tn+a; /*赋值后的tn为i个 a组成数的值*/sn=sn+tn; /*赋值后的sn为多项式前i项之和*/a=a*10;++i;}printf("a+aa+aaa+...=%d\n",sn);return 0;}5-6#include <>int main(){double s=0,t=1;int n;for (n=1;n<=20;n++){t=t*n;s=s+t;}printf("1!+2!+...+20!=%\n",s);return 0;}5-7#include <>int main(){int n1=100,n2=50,n3=10;double k,s1=0,s2=0,s3=0;for (k=1;k<=n1;k++) /*计算1到100的和*/{s1=s1+k;}for (k=1;k<=n2;k++) /*计算1到50各数的平方和*/ {s2=s2+k*k;}for (k=1;k<=n3;k++) /*计算1到10的各倒数和*/ {s3=s3+1/k;}printf("sum=%\n",s1+s2+s3);return 0;}5-8#include <>int main(){int i,j,k,n;printf("parcissus numbers are ");for (n=100;n<1000;n++){i=n/100;j=n/10-i*10;k=n%10;if (n==i*i*i + j*j*j + k*k*k)printf("%d ",n);}printf("\n");return 0;}5-9#include <>int main(){int m,s,i;for (m=2;m<1000;m++){s=0;for (i=1;i<m;i++)if ((m%i)==0) s=s+i;if(s==m){printf("%d,its factors are ",m);for (i=1;i<m;i++)if (m%i==0) printf("%d ",i);printf("\n");}}return 0;}5-10#include <>int main(){int i,n=20;double a=2,b=1,s=0,t;for (i=1;i<=n;i++){s=s+a/b;t=a,a=a+b,b=t;}printf("sum=%\n",s);return 0;}5-11#include <>int main(){double sn=100,hn=sn/2;int n;for (n=2;n<=10;n++){sn=sn+2*hn; /*第n次落地时共经过的米数*/ hn=hn/2; /*第n次反跳高度*/}printf("第10次落地时共经过%f米\n",sn);printf("第10次反弹%f米\n",hn);return 0;}5-12#include <>int main(){int day,x1,x2;day=9;x2=1;while(day>0){x1=(x2+1)*2; /*第1天的桃子数是第2天桃子数加1后的2倍.*/ x2=x1;day--;}printf("total=%d\n",x1);return 0;}5-13#include <>#include <>int main(){float a,x0,x1;printf("enter a positive number:");scanf("%f",&a);x0=a/2;x1=(x0+a/x0)/2;do{x0=x1;x1=(x0+a/x0)/2;}while(fabs(x0-x1)>=1e-5);printf("The square root of % is %\n",a,x1);return 0;}5-14#include <>#include <>int main(){double x1,x0,f,f1;x1=;do{x0=x1;f=((2*x0-4)*x0+3)*x0-6;f1=(6*x0-8)*x0+3;x1=x0-f/f1;}while(fabs(x1-x0)>=1e-5);printf("The root of equation is %\n",x1);return 0;}5-15#include <>#include <>int main(){float x0,x1,x2,fx0,fx1,fx2; do{printf("enter x1 & x2:"); scanf("%f,%f",&x1,&x2); fx1=x1*((2*x1-4)*x1+3)-6; fx2=x2*((2*x2-4)*x2+3)-6; }while(fx1*fx2>0);do{x0=(x1+x2)/2;fx0=x0*((2*x0-4)*x0+3)-6; if ((fx0*fx1)<0){x2=x0;fx2=fx0;}else{x1=x0;fx1=fx0;}}while(fabs (fx0)>=1e-5); printf("x=%\n",x0);return 0;}5-16#include <>int main(){int i,j,k;for (i=0;i<=3;i++){for (j=0;j<=2-i;j++)printf(" ");for (k=0;k<=2*i;k++)printf("*");printf("\n");}for (i=0;i<=2;i++){for (j=0;j<=i;j++)printf(" ");for (k=0;k<=4-2*i;k++)printf("*");printf("\n");}return 0;}5-17#include <>int main(){char i,j,k; /*是a的对手;j是b的对手;k是c的对手*/ for (i='x';i<='z';i++)for (j='x';j<='z';j++)if (i!=j)for (k='x';k<='z';k++)if (i!=k && j!=k)if (i!='x' && k!='x' && k!='z')printf("A--%c\nB--%c\nC--%c\n",i,j,k);return 0;}6-1#include <>#include <>int main(){int i,j,n,a[101];for (i=1;i<=100;i++)a[i]=i;a[1]=0;for (i=2;i<sqrt(100);i++)for (j=i+1;j<=100;j++){if(a[i]!=0 && a[j]!=0)if (a[j]%a[i]==0)a[j]=0;}printf("\n");for (i=2,n=0;i<=100;i++){ if(a[i]!=0){printf("%5d",a[i]);n++;}if(n==10){printf("\n");n=0;}}printf("\n");return 0;}6-2#include <>int main(){int i,j,min,temp,a[11];printf("enter data:\n");for (i=1;i<=10;i++){printf("a[%d]=",i);scanf("%d",&a[i]);}printf("\n");printf("The orginal numbers:\n"); for (i=1;i<=10;i++)printf("%5d",a[i]);printf("\n");for (i=1;i<=9;i++){min=i;for (j=i+1;j<=10;j++)if (a[min]>a[j]) min=j;temp=a[i];a[i]=a[min];a[min]=temp;}printf("\nThe sorted numbers:\n"); for (i=1;i<=10;i++)printf("%5d",a[i]);printf("\n");return 0;}6-3#include <>int main(){int a[3][3],sum=0;int i,j;printf("enter data:\n");for (i=0;i<3;i++)for (j=0;j<3;j++)scanf("%3d",&a[i][j]);for (i=0;i<3;i++)sum=sum+a[i][i];printf("sum=%6d\n",sum);return 0;}6-4#include <>int main(){ int a[11]={1,4,6,9,13,16,19,28,40,100};int temp1,temp2,number,end,i,j;printf("array a:\n");for (i=0;i<10;i++)printf("%5d",a[i]);printf("\n");printf("insert data:");scanf("%d",&number);end=a[9];if (number>end)a[10]=number;else{for (i=0;i<10;i++){if (a[i]>number){temp1=a[i];a[i]=number;for (j=i+1;j<11;j++){temp2=a[j];a[j]=temp1;temp1=temp2;}break;}}}printf("Now array a:\n");for (i=0;i<11;i++)printf("%5d",a[i]);printf("\n");return 0;}6-5#include <>#define N 5int main(){ int a[N],i,temp;printf("enter array a:\n");for (i=0;i<N;i++)scanf("%d",&a[i]);printf("array a:\n");for (i=0;i<N;i++)printf("%4d",a[i]);for (i=0;i<N/2;i++) n",number);;printf("continu or not(Y/N)");scanf(" %c",&c);if (c=='N'||c=='n')flag=0;}return 0;}6-10#include <>int main(){int i,j,upp,low,dig,spa,oth;char text[3][80];upp=low=dig=spa=oth=0;for (i=0;i<3;i++){ printf("please input line %d:\n",i+1);gets(text[i]);for (j=0;j<80 && text[i][j]!='\0';j++){if (text[i][j]>='A'&& text[i][j]<='Z') upp++;else if (text[i][j]>='a' && text[i][j]<='z') low++;else if (text[i][j]>='0' && text[i][j]<='9') dig++;else if (text[i][j]==' ')spa++;elseoth++;}}printf("\nupper case: %d\n",upp);printf("lower case: %d\n",low);printf("digit : %d\n",dig);printf("space : %d\n",spa);printf("other : %d\n",oth);return 0;}6-11#include <>int main(){ char a[5]={'*','*','*','*','*'};int i,j,k;char space=' ';for (i=0;i<5;i++){ printf("\n");printf(" ");for (j=1;j<=i;j++)printf("%c",space);for (k=0;k<5;k++)printf("%c",a[k]);}printf("\n");return 0;}6-12a-c#include <>int main(){ int j,n;char ch[80],tran[80];printf("input cipher code:");gets(ch);printf("\ncipher code :%s",ch);j=0;while (ch[j]!='\0'){ if ((ch[j]>='A') && (ch[j]<='Z'))tran[j]=155-ch[j];else if ((ch[j]>='a') && (ch[j]<='z')) tran[j]=219-ch[j];elsetran[j]=ch[j];j++;}n=j;printf("\noriginal text:");for (j=0;j<n;j++)putchar(tran[j]);printf("\n");return 0;}6-12b#include <>int main(){int j,n;char ch[80];printf("input cipher code:\n");gets(ch);printf("\ncipher code:%s\n",ch);j=0;while (ch[j]!='\0'){ if ((ch[j]>='A') && (ch[j]<='Z'))ch[j]=155-ch[j];else if ((ch[j]>='a') && (ch[j]<='z'))ch[j]=219-ch[j];elsech[j]=ch[j];j++;}n=j;printf("original text:");for (j=0;j<n;j++)putchar(ch[j]);printf("\n");return 0;}6-13#include <>int main(){ char s1[80],s2[40];int i=0,j=0;printf("input string1:");scanf("%s",s1);printf("input string2:");scanf("%s",s2);while (s1[i]!='\0')i++;while(s2[j]!='\0')s1[i++]=s2[j++];s1[i]='\0';printf("\nThe new string is:%s\n",s1);return 0;}6-14#include <>int main(){ int i,resu;char s1[100],s2[100];printf("input string1:");gets(s1);printf("\ninput string2:");gets(s2);i=0;while ((s1[i]==s2[i]) && (s1[i]!='\0'))i++; if (s1[i]=='\0' && s2[i]=='\0')resu=0;elseresu=s1[i]-s2[i];printf("\nresult:%d.\n",resu); return 0;}6-15#include <>#include <>int main(){ char s1[80],s2[80];int i;printf("input s2:");scanf("%s",s2);for (i=0;i<=strlen(s2);i++)s1[i]=s2[i];printf("s1:%s\n",s1);return 0;}7-1-1#include <>int main(){int hcf(int,int);int lcd(int,int,int);int u,v,h,l;scanf("%d,%d",&u,&v);h=hcf(u,v);printf("",h);l=lcd(u,v,h);printf("",l);return 0;}int hcf(int u,int v){int t,r;if (v>u){t=u;u=v;v=t;}while ((r=u%v)!=0){u=v;v=r;}return(v);}int lcd(int u,int v,int h){return(u*v/h);}7-1-2#include <>int Hcf,Lcd;int main(){void hcf(int,int);void lcd(int,int);int u,v;scanf("%d,%d",&u,&v);hcf(u,v);lcd(u,v);printf("",Hcf);printf("",Lcd);return 0;}void hcf(int u,int v){int t,r;if (v>u){t=u;u=v;v=t;}while ((r=u%v)!=0){u=v;v=r;}Hcf=v;}void lcd(int u,int v){Lcd=u*v/Hcf;}7-2#include <>#include <>float x1,x2,disc,p,q;int main(){void greater_than_zero(float,float);void equal_to_zero(float,float);void smaller_than_zero(float,float);float a,b,c;printf("input a,b,c:");scanf("%f,%f,%f",&a,&b,&c);printf("equation: %*x*x+%*x+%=0\n",a,b,c);disc=b*b-4*a*c;printf("root:\n");if (disc>0){greater_than_zero(a,b);printf("x1=%f\t\tx2=%f\n",x1,x2);}else if (disc==0){equal_to_zero(a,b);printf("x1=%f\t\tx2=%f\n",x1,x2);}else{smaller_than_zero(a,b);printf("x1=%f+%fi\tx2=%f-%fi\n",p,q,p,q); }return 0;}void greater_than_zero(float a,float b){x1=(-b+sqrt(disc))/(2*a);x2=(-b-sqrt(disc))/(2*a);}void equal_to_zero(float a,float b){x1=x2=(-b)/(2*a);}void smaller_than_zero(float a,float b){p=-b/(2*a);q=sqrt(-disc)/(2*a);}7-3#include <>int main(){int prime(int);int n;printf("input an integer:");scanf("%d",&n);if (prime(n))printf("%d is a prime.\n",n);elseprintf("%d is not a prime.\n",n);return 0;}int prime(int n){int flag=1,i;for (i=2;i<n/2 && flag==1;i++) if (n%i==0)flag=0;return(flag);}7-4#include <>#define N 3int array[N][N];int main(){ void convert(int array[][3]); int i,j;printf("input array:\n");for (i=0;i<N;i++)for (j=0;j<N;j++)scanf("%d",&array[i][j]);printf("\noriginal array :\n"); for (i=0;i<N;i++){for (j=0;j<N;j++)printf("%5d",array[i][j]);printf("\n");}convert(array);printf("convert array:\n");for (i=0;i<N;i++){for (j=0;j<N;j++)printf("%5d",array[i][j]); printf("\n");}return 0;}void convert(int array[][3]){int i,j,t;for (i=0;i<N;i++)for (j=i+1;j<N;j++){t=array[i][j];array[i][j]=array[j][i];array[j][i]=t;}}#include <>#include <>int main(){void inverse(char str[]);char str[100];printf("input string:");scanf("%s",str);inverse(str);printf("inverse string:%s\n",str);return 0;}void inverse(char str[]){char t;int i,j;for (i=0,j=strlen(str);i<(strlen(str)/2);i++,j--){t=str[i];str[i]=str[j-1];str[j-1]=t;}}7-6#include <>int main(){void concatenate(char string1[],char string2[],char string[]); char s1[100],s2[100],s[100];printf("input string1:");scanf("%s",s1);printf("input string2:");scanf("%s",s2);concatenate(s1,s2,s);printf("\nThe new string is %s\n",s);return 0;}void concatenate(char string1[],char string2[],char string[]) {int i,j;for (i=0;string1[i]!='\0';i++)string[i]=string1[i];for(j=0;string2[j]!='\0';j++)string[i+j]=string2[j];string[i+j]='\0';}7-7#include <>int main(){void cpy(char [],char []);char str[80],c[80];printf("input string:");gets(str);cpy(str,c);printf("The vowel letters are:%s\n",c);return 0;}void cpy(char s[],char c[]){ int i,j;for (i=0,j=0;s[i]!='\0';i++)if (s[i]=='a'||s[i]=='A'||s[i]=='e'||s[i]=='E'||s[i]=='i'|| s[i]=='I'||s[i]=='o'||s[i]=='O'||s[i]=='u'||s[i]=='U'){c[j]=s[i];j++;}c[j]='\0';}7-8#include <>#include <>int main(){char str[80];void insert(char []);printf("input four digits:");scanf("%s",str);insert(str);return 0;}void insert(char str[]){int i;for (i=strlen(str);i>0;i--){str[2*i]=str[i];str[2*i-1]=' ';}printf("output:\n%s\n",str);}7-9#include <>int letter,digit,space,others;int main(){void count(char []);char text[80];printf("input string:\n");gets(text);printf("string:");puts(text);letter=0;digit=0;space=0;others=0;count(text);printf("\nletter:%d\ndigit:%d\nspace:%d\nothers:%d\n",letter,digit,space,others );return 0;}void count(char str[]){int i;for (i=0;str[i]!='\0';i++)if ((str[i]>='a'&& str[i]<='z')||(str[i]>='A' && str[i]<='Z'))letter++;else if (str[i]>='0' && str [i]<='9')digit++;else if (str[i]==32)space++;elseothers++;}7-10#include <>#include <>int main(){int alphabetic(char);int longest(char []);int i;char line[100];printf("input one line:\n");gets(line);printf("The longest word is :");for (i=longest(line);alphabetic(line[i]);i++) printf("%c",line[i]);printf("\n");return 0;}int alphabetic(char c){if ((c>='a' && c<='z')||(c>='A'&&c<='z'))return(1);elsereturn(0);}int longest(char string[]){int len=0,i,length=0,flag=1,place=0,point;for (i=0;i<=strlen(string);i++)if (alphabetic(string[i]))if (flag){point=i;flag=0;}elselen++;else{flag=1;if (len>=length){length=len;place=point;len=0;}}return(place);}7-11#include <>#include <>#define N 10char str[N];int main(){void sort(char []);int i,flag;for (flag=1;flag==1;){printf("input string:\n");scanf("%s",&str);if (strlen(str)>N)printf("string too long,input again!"); elseflag=0;}sort(str);printf("string sorted:\n");for (i=0;i<N;i++)printf("%c",str[i]);printf("\n");return 0;}void sort(char str[]){int i,j;char t;for(j=1;j<N;j++)for (i=0;(i<N-j)&&(str[i]!='\0');i++)if(str[i]>str[i+1]){t=str[i];str[i]=str[i+1];str[i+1]=t;}}7-12#include <>#include <>int main(){float solut(float a,float b,float c,float d); float a,b,c,d;printf("input a,b,c,d:");scanf("%f,%f,%f,%f",&a,&b,&c,&d);printf("x=%\n",solut(a,b,c,d));return 0;}float solut(float a,float b,float c,float d) {float x=1,x0,f,f1;do{x0=x;f=((a*x0+b)*x0+c)*x0+d;f1=(3*a*x0+2*b)*x0+c;x=x0-f/f1;}while(fabs(x-x0)>=1e-3);return(x);}7-16#include <>#define MAX 1000int main(){ int htoi(char s[]);int c,i,flag,flag1;char t[MAX];i=0;flag=0;flag1=1;printf("input a HEX number:");while((c=getchar())!='\0' && i<MAX&& flag1){if (c>='0' && c<='9'||c>='a' && c<='f'||c>='A' && c<='F') {flag=1;t[i++]=c;}else if (flag){t[i]='\0';printf("decimal number %d\n",htoi(t));printf("continue or not");c=getchar();if (c=='N'||c=='n')flag1=0;else{flag=0;i=0;printf("\ninput a HEX number:");}}}return 0;}int htoi(char s[]){ int i,n;n=0;for (i=0;s[i]!='\0';i++){if (s[i]>='0'&& s[i]<='9')n=n*16+s[i]-'0';if (s[i]>='a' && s[i]<='f')n=n*16+s[i]-'a'+10;if (s[i]>='A' && s[i]<='F')n=n*16+s[i]-'A'+10;}return(n);}7-17#include <>int main(){ void convert(int n);int number;printf("input an integer: ");scanf("%d",&number);printf("output: ");if (number<0){putchar('-');putchar(' '); /* 先输出一个‘-’号和空格 */ number=-number;}convert(number);printf("\n");return 0;}void convert(int n){ int i;if ((i=n/10)!=0)convert(i);putchar(n%10+'0');putchar(32);}7-18#include <>int main(){int sum_day(int month,int day);int leap(int year);int year,month,day,days;printf("input date(year,month,day):");scanf("%d,%d,%d",&year,&month,&day);printf("%d/%d/%d ",year,month,day);days=sum_day(month,day); /* 调用函数sum_day */ if(leap(year)&&month>=3) /* 调用函数leap */days=days+1;printf("is the %dth day in this year.\n",days);return 0;}int sum_day(int month,int day) /* 函数sum_day:计算日期 */ {int day_tab[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};int i;for (i=1;i<month;i++)day+=day_tab[i]; /* 累加所在月之前天数 */return(day);} /* 函数leap:判断是否为闰年 */int leap(int year){int leap;leap=year%4==0&&year%100!=0||year%400==0;return(leap);}8-1#include <>int main(){ void swap(int *p1,int *p2);int n1,n2,n3;int *p1,*p2,*p3;printf("input three integer n1,n2,n3:");scanf("%d,%d,%d",&n1,&n2,&n3);p1=&n1;p2=&n2;p3=&n3;if(n1>n2) swap(p1,p2);if(n1>n3) swap(p1,p3);if(n2>n3) swap(p2,p3);printf("Now,the order is:%d,%d,%d\n",n1,n2,n3);return 0;}void swap(int *p1,int *p2){int p;p=*p1; *p1=*p2; *p2=p;}8-2#include <>#include <>int main(){void swap(char *,char *);char str1[20],str2[20],str3[20];printf("input three line:\n");gets(str1);gets(str2);gets(str3);if(strcmp(str1,str2)>0) swap(str1,str2); if(strcmp(str1,str3)>0) swap(str1,str3); if(strcmp(str2,str3)>0) swap(str2,str3); printf("Now,the order is:\n");printf("%s\n%s\n%s\n",str1,str2,str3);return 0;}void swap(char *p1,char *p2){char p[20];strcpy(p,p1);strcpy(p1,p2);strcpy(p2,p); }8-3#include <>int main(){ void input(int *);void max_min_value(int *);void output(int *);int number[10];input(number); max_min_value(number); output(number);return 0;}void input(int *number){int i;printf("input 10 numbers:");for (i=0;i<10;i++)scanf("%d",&number[i]);}void max_min_value(int *number) { int *max,*min,*p,temp;max=min=number;for (p=number+1;p<number+10;p++)if (*p>*max) max=p;else if (*p<*min) min=p;。

C语言程序设计实验报告学校：内蒙古科技大学经济与管理学院班级：信息管理与信息系统2011级姓名：王*学号：**********目录实验一C语言开发环境建立 (3)实验二C语言程序初步 (3)实验三变量与赋值 (4)实验四运算符与表达式 (5)实验五数据类型 (7)实验六控制台输入/输出 (9)实验七关系运算与逻辑运算 (11)实验八分支程序的测试与调试 (12)实验九条件型循环结构及其测试 (16)实验十计数型循环结构及其测试 (20)实验十一函数及其测试 (24)实验十二变量的储存性质 (26)实验十三数组 (30)实验十四指针 (32)实验十五结构体和共用体 (37)实验十六文件 (40)实验11、实验目的与要求（1）掌握C语言开发环境的软/硬件配置和安装方法。

（2）为了自己的方便进行工作环境配置。

（3）测试所建立的C语言程序开发环境。

2、实验测试一个简单的程序，对所建立的程序开发环境进行测试例如，可以准备一个简单的程序如下。

#include <stdio.h>int main(void){printf("I am studying programming with C.");}编译结果运行结果3.分析与预期结果相同实验21.实验目的与要求（1）进一步熟悉c语言程序设计环境。

（2）明确c语言程序的结构。

2. 程序检测与调试程序一错误程序：main()(printf("This is c program.");)调试结果：程序缺少“#include<stdio.h>”，main函数下面应该是“｛｝”而不是“（）”。

改正后的程序：#include<stdio.h>main(){printf("This is c program.\n");}程序二错误程序：#include<stdio.h>;main(){a=3,b=5;c=a*b;printf("The product is:",c)}调试结果：“#include<stdio.h>”后面没有“；”，对变量a,b,c没有定义，“｝”前缺少“；”。