}
printf("%d is the biggest number !\n",s);
return 0;
}
P036 按大小顺序输出一些数.
#include<>
int main()
{
int i , j , a[4] , s=0 ;
printf("Please input 5 numbers :\n"); for (i=0 ; i<=4 ; i++) {
scanf("%d",&a[i]); }
for (i=0 ; i<=3 ; i++) {
for (j=i+1 ; j<=4 ; j++) {
if (a[i]>a[j]) {
s=a[i];
a[i]=a[j];
a[j]=s;
}
}
}
for (i=0 ; i<=4 ; i++) printf("%d-",a[i]);
return 0;
}
P036 求1至100的总合.
#include<>
int main()
{
int i , sum=0 ; for (i=0 ; i<101 ; i++)
sum=sum+i;
printf("The sum of one to one hundred is %d !\n",sum);
return 0;
}
P036 判断一个数能否同时被3和5整除.
#include<>
int main()
{
int n ;
printf("Please input a number :\n");
scanf("%d",&n);
if(n%3==0&&n%5==0) printf ("Can be devide by 3 and 5 !\n");
else
printf ("Can not be devide by 3 and 5 !\n");
return 0;
}
#include<> #include<>qrt是求根,属数学函数.
int main() {
int i;
for (i=100; i<=200; i++) if(prime_number(i) == 1)
printf("%d ",i);
return 0;
}
int prime_number(double m) {
int j,k;
k=(int)sqrt(m); for(j=2;j<=k;j++)
{
if(m%j==0)
return 0; }
return 1;
}
#include<>
int main()
{
int i;
for(i=100;i<=200;i++) {
if(prime(i)==1) printf ("%d is the prime number !\n",i);
}
return 0;
}
int prime(int j) {
int m, n;
m=(int)sqrt(j);
for (n=2;n<=m;n++)
{
if(j%n==0)
return 0; }
return 1;
}
请仿照来写.
P036 最大公约数和最小公倍数.
#include<> main ()
{
int m, n, c, d;
int gcd(); int lcm(); printf("Please input two number :\n");
scanf("%d %d",&m,&n);
c=gcd(m,n); y=x%y; x=temp; }
return y; }
int lcm(int x, int y) 于号降序,大于号升序.
{ temp=x;
x=y;
y=temp;
}
for(i=1; i<=y; i++) {
if(!((x*i)%y)) { return x*i;
}
}
}
最简单的C程序设计——顺序程序设计
float F_to_C(float input_fah) {
float output_cen; output_cen=9)*(input_fah-32); return
output_cen; }
float C_to_F(float input_cen)
{
float output_fah;
output_fah=5)*input_cen+32; return output_fah;
}
int main()
{
int choice;
float input_fah,input_cen,output_fah,output_cen; printf("F_to_C press <1> and C_to_F press <2> !\n");
scanf("%d",&choice);
if(choice==1)
{
printf("Please input fahrenheit :");
scanf("%f",&input_fah); output_cen=F_to_C(input_fah);
printf("The 华氏 is %d , 摄氏 is %d .",(int)input_fah,(int)output_cen);
}
if(choice==2)
{
printf("Please input centigrade :");
scanf("%f",&input_cen);
output_fah=C_to_F(input_cen);
printf("The Centigrade is %d , and the Fahrenheit
is %d .",(int)input_cen,(int)output_fah);
}
return 0;
}
P038 计算存款利息(关于精度问题).
#include<>
int main()
{
float p0=1000,r1=,r2=,r3=,p1,p2,p3;
p1=p0*(1+r1);
p2=p0*(1+r2);
p3=p0*(1+r3/2)*(1+r3/2);
printf("p1=%f\np2=%f\np3=%f\n",p1,p2,p3);
return 0;
}
P055 大写转换成小写
#include<>
int main() 6个字母.
{
char c1, c2;
c1='A';
c2=c1+32;
printf("%c %d",c2,c2);
return 0;
}
P059 给出三角形边长,算出面积.
#include<>
#include<>
int main()
{
double a=, b=, c=, s, area;
s=(a+b+c)/2;
area=sqrt(s*(s-a)*(s-b)*(s-c));
printf("area is %f\n",area); return 0;
}
P065 求一无二次等式的根,默认两个不同根.
#include<>
#include<>
int main()
{
double a,b,c,disc,x1,x2,p,q;
scanf("%lf %lf %lf",&a,&b,&c);
disc=b*b-4*a*c;
p=-b/*a);
q=sqrt(disc)/*a);
x1=p+q;
x2=p-q;
printf("x1=%\nx2=%",x1,x2);
return 0;
}
#include<>
#include<>
int main()
{
double a=; 以是float.
printf("%.9f\n",a/3);
return 0;
}
#include<>
#include<>
int main()
{
float a; 33252,float精度6位,所以第七位后不可信.
a=10000/;
printf("%f\n",a);
return 0;
}
P078 使用putchar函数输出.
#include<>
#include<>
int main()
{
char a='B',b='O',c='Y';
putchar(a);
putchar(b);
putchar(c);
putchar('\n');
putchar(101); putchar(66);
return 0;
}
#include<>
#include<>
int main()
{
char a,b,c;
a=getchar();
b=getchar();
c=getchar();
putchar(a);
putchar(b);
putchar(c); putchar('\n');
return 0;
}
#include<>
#include<>
int main()
{
char a,b;
a=getchar();
b=a+32;
putchar(b);
putchar('\n');
return 0;
}
P082 国民生产总值10年后的增长倍数.
#include<>
#include<>
int main()
{
double p,r=,n=10;
p=pow((1+r),n); printf("P is %lf when 10 years later .\n",p);
return 0; }
P082 求各种存款的利息数.
#include<>
#include<>
int main()
{
double p,r,n; p=1000*(1+5*;
printf("5 years is %lf !\n",p); f输出的是double型.
p=(1000*(1+2*);
p=(p*(1+3*);
printf("5 years is %lf !\n",p); p=(1000*(1+3*);
p=(p*(1+2*);
printf("5 years is %lf !\n",p); 明,是一样的.
p=1000*pow((1+,5);
printf("5 years is %lf !\n",p); p=1000*pow((1+4),4*5);
printf("5 years is %lf !\n",p); #include<>
#include<>
int main()
{
double m,r=,d=300000,p=6000;
m=(log10(p/(p-d*r)))/(log10(1+r));
printf("%.1lf",m); lf.
return 0;
}
P084 字母密码转换,调用函数及临界处理.
#include<>
char printcode(char f)
{
if(((int)f>86&&(int)f<91)||((int)f>118&&(int)f<123))
{
return(f-26+4); }
else
{
return(f+4);
}
}
int main()
{
char a,b,c,d,e;
printf("Please input :\n");
a=getchar();
b=getchar();
c=getchar();
d=getchar();
e=getchar();
printf("%c%c%c%c%c",printcode(a),printcode(b),printcode(c),printcode(d),printcode(e) );
putchar(putcharcode(a));
putchar(putcharcode(b));
putchar(putcharcode(c));
putchar(putcharcode(d));
putchar(putcharcode(e));
return 0; lf 来实现,因为没有要求实部,所以格式中m不写.
以转换,但要在某此条件下,例如输出和读入时,%c是字母,而%d是数值,看着办. }
选择结构程序设计
P086 一无二次方程求根的二分支.
#include<>
#include<>
int main()
{
double a,b,c,disc,x1,x2,p,q;
scanf("%lf %lf %lf",&a,&b,&c);
disc=b*b-4*a*c;
if(disc<0) printf("This equation hasn't real roots\n");
else
{
p=-b/*a);
q=sqrt(disc)/*a);
x1=p+q;
x2=p-q;
printf("x1=%\nx2=%",x1,x2);
}
return 0;
}
P087 二个数按大小输出.
#include<>
int main() {
float a,b,t;
scanf("%f %f",&a,&b); 如有个逗号.
if(a>b)
{
t=a;
a=b;
b=t;
}
printf("%,%\n",a,b);
return 0;
}
P088 三个数按大小输出.
#include<>
int main() {
float a,b,c,t;
scanf("%f %f %f",&a,&b,&c);
if(a>b) {
t=a;
a=b;
b=t;
}
if(a>c) {
t=a;
a=c;
c=t;
}
if(b>c) {
t=b;
b=c;
c=t;
}
printf("%,%%\n",a,b,c);
return 0;
}
P099 判断输入字符,并最终按小写输出.
#include<>
int main()
{
char ch;
scanf("%c",&ch);
ch=(ch>='A'&&ch<='Z')(ch+32):ch; printf("%c\n",ch);
return 0;
}
P100 按要求输出相应的Y值.
#include<>
int main()
{
int x,y;
scanf("%d",&x);
if(x>=0)
{
if(x>0) {
y=1;
}
else
{
y=0;
}
}
else{
y=-1;
}
printf("x=%d,y=%d",x,y);
return 0;
}
P102 switch的简单应用.
#include<>
int main()
{
char grade;
scanf("%c",&grade);
printf("Your score:");
switch(grade)
{
case'a':printf("85-100\n");break; case'b':printf("70-84\n");break;
case'c':printf("60-69\n");break;
case'd':printf("<60\n");break;
default:printf("Enter data error!\n");
}
return 0;
}
#include<>
void action1(int x,int y)
{
printf("x+y=%d\n",x+y);
}
void action2(int x,int y)
{
printf("x*y=%d\n",x*y);
}
int main()
{
char ch;
int a=15,b=23;
ch=getchar();
switch(ch)
{
case'a':
case'A':action1(a,b);break; case'b':
case'B':action2(a,b);break;
default:putchar('\a'); }
return 0;
}
P106 用if的分支来做闰年问题
#include<>
int main()
{
int year,leap;
printf("Please input the year:\n");
scanf("%d",&year);
if(year%4==0) {
if(year%100==0) {
if(year%400==0) {
leap=1;
}
else{
leap=0;
}
}
else}
}
else
{
leap=0;
}
if(leap)
{
printf("%d is ",year);
}
else
{
printf("%d is not ",year);
}
printf("a leap year !"); return 0;
}
P108 一元二次等式的全计算过程.
#include<>
#include<>
int main()
{
double a,b,c,disc,x1,x2,realpart,imagpart;
scanf("%lf %lf %lf",&a,&b,&c);
printf("The equation ");
if(fabs(a)<=1e-6) {
printf("is not a quadratic !\n");
printf("x1=x2=%lf",-c/b);
}
else
{
disc=b*b-4*a*c;
if(fabs(disc)<=1e-6) {
printf("has two equal roots : %lf\n",-b/(2*a));
}
else
{
if(disc>1e-6)
{
x1=(-b+sqrt(disc))/(2*a);
x2=(-b-sqrt(disc))/(2*a);
printf("has distinct real roots : %lf and %lf \n",x1,x2);
}
else
{
realpart=-b/(2*a);
imagpart=sqrt(-disc)/(2*a);
printf("has complex roots: \n");
printf("%lf + %lfi\n",realpart,imagpart);
printf("%lf + %lfi\n",realpart,imagpart);
}
}
}
return 0;