C++程序设计习题答案第六章

第六章课后习题答案（第二版谭浩强）1://xt6-1/cpp#include <iostream> //如用VC++应改为∶#include <iosttram.h>using namespace std; //如用VC++应取消此行#include "cylinder.h"#include "point.cpp"#include "circle.cpp"#include "cylinder.cpp"int main(){Cylinder cy1(3.5,6.4,5.2,10);cout<<"\noriginal cylinder:\nx="<<cy1.getX()<<", y="<<cy1.getY()<<", r="<<cy1.getRadius()<<",h="<<cy1.getHeight()<<"\narea="<<cy1.area()<<", volume="<<cy1.volume()<<endl;cy1.setHeight(15);cy1.setRadius(7.5);cy1.setPoint(5,5);cout<<"\nnew cylinder:\n"<<cy1;Point &pRef=cy1;cout<<"\npRef as a point:"<<pRef;Circle &cRef=cy1;cout<<"\ncRef as a Circle:"<<cRef;return 0;}3:解法一#include <iostream>using namespace std;class Point{public:Point(float a,float b):x(a),y(b){}~Point(){cout<<"executing Point destructor"<<endl;}private:float x;float y;};class Circle:public Point{public:Circle(float a,float b,float r):Point(a,b),radius(r){} ~Circle(){cout<<"executing Circle destructor"<<endl;} private:float radius;};int main(){Point *p=new Circle(2.5,1.8,4.5);delete p;return 0;}3：解法二#include <iostream>using namespace std;class Point{public:Point(float a,float b):x(a),y(b){}~Point(){cout<<"executing Point destructor"<<endl;} private:float x;float y;};class Circle:public Point{public:Circle(int a,int b,int r):Point(a,b),radius(r){}~Circle(){cout<<"executing Circle destructor"<<endl;} private:float radius;};int main(){Point *p=new Circle(2.5,1.8,4.5);Circle *pt=new Circle(2.5,1.8,4.5);delete pt;return 0;}3：解法三#include <iostream>using namespace std;class Point{public:Point(float a,float b):x(a),y(b){}virtual ~Point(){cout<<"executing Point destructor"<<endl;}private:float x;float y;};class Circle:public Point{public:Circle(float a,float b,float r):Point(a,b),radius(r){}virtual ~Circle(){cout<<"executing Circle destructor"<<endl;}private:float radius;};void main(){Point *p=new Circle(2.5,1.8,4.5);delete p;}4：#include <iostream>using namespace std;//定义抽象基类Shapeclass Shape{public:virtual double area() const =0; //纯虚函数};//定义Circle类class Circle:public Shape{public:Circle(double r):radius(r){} //结构函数virtual double area() const {return 3.14159*radius*radius;};//定义虚函数protected:double radius; //半径};//定义Rectangle类class Rectangle:public Shape{public:Rectangle(double w,double h):width(w),height(h){} //结构函数virtual double area() const {return width*height;} //定义虚函数protected:double width,height; //宽与高};class Triangle:public Shape{public:Triangle(double w,double h):width(w),height(h){} //结构函数virtual double area() const {return 0.5*width*height;}//定义虚函数protected:double width,height; //宽与高};//输出面积的函数void printArea(const Shape &s){cout<<s.area()<<endl;}//输出s的面积int main(){Circle circle(12.6); //建立Circle类对象circlecout<<"area of circle =";printArea(circle);//输出circle的面积Rectangle rectangle(4.5,8.4); //建立Rectangle类对象rectanglecout<<"area of rectangle =";printArea(rectangle);//输出rectangle的面积Triangle triangle(4.5,8.4); //建立Triangle类对象cout<<"area of triangle =";printArea(triangle); //输出triangle的面积return 0;}5：#include <iostream>using namespace std;//定义抽象基类Shapeclass Shape{public:virtual double area() const =0; //纯虚函数};//定义Circle(圆形)类class Circle:public Shape{public:Circle(double r):radius(r){}//结构函数virtual double area() const {return 3.14159*radius*radius;};//定义虚函数protected:double radius; //半径};//定义Square(正方形)类class Square:public Shape{public:Square(double s):side(s){} //结构函数virtual double area() const {return side*side;} //定义虚函数protected:double side;};//定义Rectangle(矩形)类class Rectangle:public Shape{public:Rectangle(double w,double h):width(w),height(h){} //结构函数virtual double area() const {return width*height;} //定义虚函数protected:double width,height; //宽与高};//定义Trapezoid(梯形)类class Trapezoid:public Shape{public:Trapezoid(double t,double b,doubleh):top(t),bottom(t),height(h){} //结构函数virtual double area() const {return0.5*(top+bottom)*height;} //定义虚函数protected:double top,bottom,height; //上底、下底与高};//定义Triangle(三角形)类class Triangle:public Shape{public:Triangle(double w,double h):width(w),height(h){} //结构函数virtual double area()const {return 0.5*width*height;}//定义虚函数protected:double width,height; //宽与高};int main(){Circle circle(12.6); //建立Circle类对象circleSquare square(3.5); //建立Square类对象squareRectangle rectangle(4.5,8.4); //建立Rectangle类对象rectangleTrapezoid trapezoid(2.0,4.5,3.2); //建立Trapezoid类对象trapezoidTriangle triangle(4.5,8.4); //建立Triangle类对象Shape*pt[5]={&circle,&square,&rectangle,&trapezoid,&triangle};//定义基类指针数组pt，使它每一个元素指向一个派生类对象double areas=0.0; //areas为总面积for(int i=0;i<5;i++){areas=areas+pt[i]->area();}cout<<"totol of all areas="<<areas<<endl; //输出总面积return 0;}（学习的目的是增长知识，提高能力，相信一分耕耘一分收获，努力就一定可以获得应有的回报）。

实验六数组练习6.2代码如下：#include <iostream>using namespace std;int main(){int num, max = 0, count = 1;//user input 6 numbersfor (int i = 0; i < 6; i++){cout << "Enter a number: ";cin >> num;//if number entered > max, max = num, count = 1 again.if (num > max){max = num;count = 1;}else if (num == max){count++;}}//displaycout << "The largest number is " << max <<endl;cout << "The largest number appears " << count << " times\n";return 0;}练习6.4代码如下：#include <iostream>using namespace std;int main (){int score[40];int upcount = 0, downcount = 0, equalcount = 0, count = 0,i = 0, j = 0, sum = 0;//input the arrays, and when user enter a minus, stop inputingdo{cout << "Please enter less than 40 students' score, if you want to stop entering you can enter a minus.\n";cout << "You have entered " << count << " scores\n";cin >> j;if ((j >= 0) && (j <= 100)){score[i] = j;j++;count++;i++;}else if (j > 100){cout << "You should enter a correct score";}}while (j >= 0);//compute the sum of the arraysfor (i = 0; i < count; i++){sum += score[i];}//compute the averageint average = sum / count;for (i = 0; i < count; i++){if (score[i] > average){upcount++;}else if (score[i] == average){equalcount++;}else if (score[i] < average){downcount++;}}//display the resultcout << "The summer scores among " << count << " students is " << sum << endl;cout << "The average scores among " << count << " students is "<< average << endl;cout << "There are(is) " << upcount << " student's(s') scores are(is) higher than average.\n";cout << "There are(is) " << equalcount << " student's(s') scores are(is) equal to average.\n";cout << "There are(is) " << downcount << " student's(s') scores are(is) lower than average.\n";return 0;}练习6.6代码如下：#include <iostream>#include <iomanip>using namespace std;bool isPrime (int num);int main (){int prime[50];for (int i = 0, num = 2; i < 50; num++){if ( isPrime(num) ){prime[i] = num;i++;}}for (int i = 0, count = 0; i < 50; i++){cout << setw(6) << prime[i];count++;if (10 == count){cout << endl;count = 0;}}system("pause");return 0;}//this function called isPrime is used to judge an integer is prime or not... bool isPrime (int num){bool isPrime = true;for (int i = 2; i <= sqrt( (double)num ); i++){if (0 == num %i){isPrime = false;break;}}return isPrime;}练习6.8代码如下：#include <iostream>using namespace std;int average(int array[], int size);double average(double array[], int size);int main (){int array1[6] = {1,2,3,4,5,6};double array2[7] = {6.0,4.4,1.9,2.9,3,4,3.5};//display the resultcout << average(array1,6) << endl << average(array2,7) << endl;return 0;}//the kind of integerint average(int array[], int size){double sum = 0;for (int i = 0; i < size; i++){sum += array[i];}return sum / size;}//the kind of integerdouble average(double array[], int size) {double sum = 0;for (int i = 0; i < size; i++){sum += array[i];}return sum / size;}练习6.10代码如下：#include <iostream>using namespace std;void min (int array[], int size);int main (){int array[8] = {2,2,4,5,10,100,2,2};min(array,8);return 0;}//打印最小元的下标void min (int array[], int size){int min = array[0];int xiabiao[10];int j = 0;for (int i = 0; i < size; i++){if ( array[i] <= min ){min = array[i];xiabiao[j] = i;j++;}}int k = j;j = 0;//保存最小元下标的个数cout << "The min of the array(s) is " << min << endl;cout << "The subscript of the minnest number is(are) ";for (; j < k; j++){cout << xiabiao[j] << endl;}}练习6.12代码如下：#include <iostream>using namespace std;void reverse (int soure[], int size);//swapvoid reverse (int soure[], int size){for (int i = 0; i < (size / 2); i++){swap ( soure[i], soure[size-1-i] );}}//check the void reverse is right or notint main (){int array[6] = {1,2,3,4,5,6};reverse (array, 6);for (int i = 0; i < 6; i++){cout << array[i] << endl;}return 0;}练习6.14代码如下：#include <iostream>#include <ctime>using namespace std;int main (){int num[100000];srand ( time(0) );for (long i = 0; i < 100000; i++){num[i] = rand();}//生成关键字int key = rand();cout << "关键字是" << key << endl;//开始计时long startTime1 = time(0);for (int i = 0; i < 100000; i++){if (key == num[i]){cout <<"关键字的下标是"<< i <<endl;}}long endTime1 = time(0);long time1 = endTime1 - startTime1;cout << "顺序搜索时间是" << time1 <<"秒\n\n";//二分搜索int low = 0;int high = 99999;//对数组进行排序,现在开始第二次计时long startTime2 = time(0);for (long i = 9999;i >= 1; i--){int xiabiao = 0, lagest = num[0];for (long j = 1;j <= i; j++){if ( num[j] > lagest ){lagest = num[j];xiabiao = j;}}//swapif ( xiabiao != i){num[xiabiao] = num[i];num[i] = lagest;}}while (high >= low){int mid = (low + high) / 2;if (key < num[mid]){high = mid - 1;}else if (key == num[mid]){cout << "关键字下标是" << mid << endl;break;}else{low = mid + 1;}}long endTime2 = time(0);long time2 = endTime2 - startTime2;cout << "排序和二分搜索花费时间是" << time2 << "秒\n\n";system("pause");return 0;}练习6.16代码如下：#include <iostream>using namespace std;void turn ( double arrays[], int size);int main (){double arrays[7] = {6.0,4.4,1.9,2.9,3.4,2.9,3.5};turn(arrays, 7);for (int i = 0; i < 7; i++){cout << arrays[i] << " ";}system("pause");return 0;}//起泡排序void turn ( double arrays[], int size){bool changed = true;do{changed = false;for (int i = 0; i < size - 1; i++){if (arrays[i] > arrays[i+1]){swap(arrays[i], arrays[i+1]);changed = true;}}} while (changed);}练习6.18代码如下：#include <iostream>using namespace std;int main (){int arrays[4][4] = {{1,2,4,5},{6,7,8,9},{10,11,12,13},{14,15,16,17}};int sum = 0;for (int i = 0, j = 0; i < 4; i++,j++){sum += arrays[i][j];}cout << "主对角线之和为" << sum << endl;system("pause");return 0;}练习6.20代码如下：#include <iostream>using namespace std;void turn ( int arrays[], int size);int main(){int time[8][7] = {{2,4,3,4,5,8,8},{7,3,4,3,3,4,4},{3,3,4,3,3,2,2},{9,3,4,7,3,4,1},{3,5,4,3,6,3,8},{3,4,4,6,3,4,4},{3,7,4,8,3,8,4},{6,3,5,9,2,7,9}};int sumofRow[8] = {0,0,0};for (int row = 0; row < 8; row++){for (int i = 0; i < 7; i++){sumofRow[row] += time[row][i];}}turn(sumofRow, 8);for (int i = 7; i >= 0; i--){cout << sumofRow[i] << endl;}system("pause");return 0;}//排序void turn ( int arrays[], int size){bool changed = true;do{changed = false;for (int i = 0; i < size - 1; i++){if (arrays[i] > arrays[i+1]){swap(arrays[i], arrays[i+1]);changed = true;}}} while (changed);}练习6.22代码如下：#include <iostream>using namespace std;//compute the sumvoid mutiplyMatrix (int a[][5],int b[][5],int c[][5],int rowsize){for (int i = 0; i < rowsize; i++){for (int j = 0; j < rowsize; j++){for (int k = 0; k < rowsize; k++){c[i][j] += (a[i][k] + b[k][j]);}}}}int main (){int a[5][5] = {{1,2,3,4,5},{1,2,3,4,5},{1,2,3,4,5},{1,2,3,4,5},{1,2,3,4,5}};int b[5][5] = {{1,2,3,4,5},{1,2,3,4,5},{1,2,3,4,5},{1,2,3,4,5},{1,2,3,4,5}};int c[5][5] = {{0,0,0,0,0},{0},{0},{0},{0}};mutiplyMatrix(a,b,c,5);for (int i = 0; i < 5; i++){for (int j = 0; j < 5; j++){cout << c[i][j] << endl;}}system("pause");return 0;}练习6.24代码如下：#include <iostream>#include <ctime>using namespace std;int main(){srand( time(0) );int chess[8][8];//随机输入0和1，并输出8*8列表for (int i = 0; i < 8; i++){for (int j = 0; j < 8; j++){chess[i][j] = rand() % 2;cout << chess[i][j];}cout << endl;}//计算每行的和for (int row = 0; row < 8; row++){int sumofrow = 0;for (int column = 0; column < 8; column++){sumofrow += chess[row][column];}if (0 == sumofrow){cout << "All 0s on row" << row + 1 << endl;}else if (8 == sumofrow){cout << "All 1s on row" << row + 1 << endl;}}//计算每列的和for (int column = 0; column < 8; column++){int sumofcolumn = 0;for (int row = 0; row < 8; row++){sumofcolumn += chess[row][column];}if (0 == sumofcolumn){cout << "All 0s on column" << column + 1 << endl;}else if (8 == sumofcolumn){cout << "All 1s on column" << column + 1 << endl;}}//计算两个对角线的和int sumofsubdiagonal1 = 0, sumofsubdiagonal2 = 0;for (int row = 0, column = 0; row < 8; row++,column++){sumofsubdiagonal1 += chess[row][7 - column];sumofsubdiagonal2 += chess[row][column];}if ( 0 == sumofsubdiagonal1 ){cout << "All 0s on subdiagonal1" << endl;}else if ( 8 == sumofsubdiagonal1 ){cout << "All 1s on subdiagonal1" << endl;}if ( 0 == sumofsubdiagonal2 ){cout << "All 0s on subdiagonal2" << endl;}else if ( 8 == sumofsubdiagonal2 ){cout << "All 1s on subdiagonal2" << endl;}system("pause");return 0;}练习6.26代码如下：#include <iostream>using namespace std;int factors(int num, int table[][2]){/*从i=2开始除，若不能被i整除则i++；若能被i整除则输出i，且num变成num除以i的商，重新把2赋值给i，循环。

第六章习题答案一、选择填空1、A2、C3、D4、B5、D6、A7、C8、A9、D 10、A 11、C 12、A13、B 14、C 15、C 16、D 17、B 18、C 19、A 20、D21、C 22、B二、判断下列描述的正确性，对者划√，错者划×。

1、√2、×3、×4、×5、√6、√7、×8、√9、× 10、√11、√ 12、√ 13、√ 14、√ 15、× 16、√ 17、√ 18、√ 19、√ 20、×21、× 22、×三、分析下列程序的输出结果。

1、运行该程序输出结果如下所示。

Default constructor calledConstructor calleda=0,b=0a=4,b=82、运行该程序输出结果如下所示。

a=7,b=93、运行该程序输出结果如下所示。

1044、运行该程序输出结果如下所示。

1035，789.5045、运行该程序输出结果如下所示。

1{}{0，1，2，3，4，5，6，7，8}1{11，12，13，14，15，16，17，18，19}{19，18，17，16，15，14，13，12，11}6、运行该程序输出结果如下所示。

Starting1:Default constructor called.Default constructor called.Default constructor called.Eding1:Starting2:Constructor: a=5,b=6Constructor: a=7,b=8Constructor: a=9,b=10Ending2:Destructor called.a=9,b=10Destructor called.a=7,b=8Destructor called.a=5,b=6Destructor called.a=5,b=6Destructor called.a=3,b=4Destructor called.a=1,b=27、运行该程序输出结果如下所示。

1.以下对一维整型数组a的定义，正确的是_。

(2分)A.int a(10) ;B.int n = 10 , a[n] ;C.int n ;scanf( "%d" , &n ) ;int a[n] ;D.int a[10] ;2.若有定义：int a[10] ;，则对a数组元素的正确引用是_。

(2分)A.a[10]B.a[3.5]C.a(5)D.a[10-10]3.对定义int a[10] = {6 , 7 , 8 , 9 , 10} ; 的正确理解是_。

(2分)A.将5个初值依次赋给a[1]--a[5]B.将5个初值依次赋给a[0]--a[4]C.将5个初值依次赋给a[6]--a[10]D.因为数组长度与初值个数不相同，所以此语句不正确4..若有定义：int a[3][4]; , 则对a数组元素的正确引用是_。

(2分)A.a[3][4]B.a[1,3]C.a[1+1][0]D.a(2)(1)5.以下对二维数组a初始化正确的语句是_。

(2分)A.int a[2][ ]={{0 , 1 , 2}, {3 , 4 , 5}};B.int a[ ][3]={{0, 1, 2}, {3, 4, 5}};C.int a[2][4]={{0, 1 , 2}, {3 , 4}, {5}};D.int a[ ][3]={{0, 1, 2}, { }, {3, 4}};6.对二维数组a进行如下初始化：int a[ ][3]={0 , 1 , 2 , 3 , 4 , 5};则a[1][1]的值是_。

(2分)A.0B.3C.4D.17.下面程序段的运行结果是_。

(2分)#include<stdio.h>int main( ){int i , x[3][3] = {1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9} ;for( i = 0 ; i < 3 ; i++ )printf( "%2d" , x[i][2-i] ) ;return 0 ;}A.1 5 9B.1 4 7C.3 5 7D.3 6 98.以下对数组s的初始化，错误的是_。

习题及实验解答第六章一、选择题1、应为4（无正确选项）2、D3、B C4、D5、A6、A7、C8、B9、A 10、A二、程序填空1、stu[i].score k=j p[k]=p[i] p[i]=temp2、p!=NULL p->data p->next3、struct node * s->data=ch r=s NULL三、编程题1、#include "iostream.h"struct staff{char num[6];char name[8];float salary[3]; //分项工资float gs; //实得工资}s[100];void main( ){int i,j,n;cin>>n; /*输入职工人数*/for(i=0;i<n;i++){cin>>s[i].num>>s[i].name;for(j=0;j<3;j++)cin>>s[i].salary[j];}cout<<"NO. salary\n";for(i=0;i<n;i++){s[i].gs= s[i].salary[0]+ s[i].salary[1]- s[i].salary[2] ;cout<<s[i].num<<'\t'<<s[i].gs<<endl;}}2、#include "iostream.h"#include "stdio.h"struct str{char ch;struct str *next;};void main(){char s[100];int i=0,num=0;struct str *insert,*head=NULL,*p;gets(s);while(s[i]!='\0'){insert=new str;insert->ch=s[i];if(head==NULL){head=insert ;head->next=NULL;}else{insert->next=head ;head=insert;}i++;}p=head;while(p!=NULL){if(p->ch>='A'&&p->ch<='Z')num++;cout<<p->ch;p=p->next;}cout<<endl<<"num="<<num<<endl; }3、#include "iostream.h"#include "stdio.h"struct node{char ch;int count;struct node *next;};void main( ){struct node *head,*p1,*p2,*insert;char c;head=NULL;while((c=getchar())!='\n'){p1=head;while(p1!=NULL&&c>p1->ch){p2=p1;p1=p1->next;}if(p1==NULL){insert=new node;insert->ch=c;insert->count=1;if(head==NULL){insert->next=head;head=insert;}else{insert->next=NULL;p2->next=insert;}}else if(c==p1->ch)p1->count++;else{insert=new node;insert->ch=c;insert->count=1;if(p1==head){insert->next=head;head=insert;}else{insert->next=p1;p2->next=insert;}}}p1=head;while(p1!=NULL){cout<<p1->ch<<' '<<p1->count<<'\t';p1=p1->next;}}4、#include "stdio.h"#include "iostream.h"struct node{int coef;int expn;struct node *next;};struct node *creat(){struct node *head,*tail,*p;int c,e;head=NULL;cin>>c>>e;while(c!=0) //约定以输入系数为0作为多项式的结束{p=new node;p->coef=c;p->expn=e;if(head==NULL)head=p;elsetail->next=p;tail=p;cin>>c>>e;}tail->next=NULL;return head;}void print(struct node *h){struct node *p;p=h;while(p!=NULL){if(p->next!=NULL)cout<<p->coef<<'x'<<p->expn<<'+';else{if(p->coef!=0&&p->expn!=0)cout<<p->coef<<'x'<<p->expn<<endl;else if(p->coef!=0)cout<<p->coef<<endl;elsecout<<endl;}p=p->next;}}void main(){struct node *h1,*p1,*h2,*p2,*h3,*p3,*newnode,*t3;h1=creat();print(h1);h2=creat();print(h2);p2=h2;h3=NULL;p1=h1;p2=h2;p3=h3;t3=h3;while(p1!=NULL&&p2!=NULL){newnode=new node;if(p1->expn>p2->expn){newnode->coef=p1->coef;newnode->expn=p1->expn;p1=p1->next;}else if(p1->expn<p2->expn){newnode->coef=p2->coef;newnode->expn=p2->expn;p2=p2->next;}else{newnode->coef=p2->coef+p1->coef;newnode->expn=p2->expn;p2=p2->next;p1=p1->next;}if(h3==NULL){h3=newnode;t3=newnode;}elset3->next=newnode;t3=newnode;}if(p1==NULL)t3->next=p2;elset3->next=p1;print(h3);}第七章一、选择题1、D2、B3、B4、A5、B6、A7、D8、B9、C 10、B二、程序填空1、fname, “w”(ch=getchar())!=’#’count++2、(c=fgetc(fp)) length++ length=03、”wb”&emp, sizeof(employer),1,fp fclose(fp) “rb”&emp, sizeof(employer),1,fp三、编程题1、#include "iostream.h"#include "stdlib.h"#include "stdio.h"void main(){FILE *fp1,*fp2;char ch;fp1=fopen("f1.txt","a");if(fp1==NULL){cout<<"can't open f1.\n";exit(1);}if((fp2=fopen("f2.txt","r"))==NULL){cout<<"can't open f2.\n";exit(1);}while(1){ch=fgetc(fp2);if(feof(fp2))break;cout<<ch;fputc(ch,fp1);}fclose(fp1);fclose(fp2);}2、#include "iostream.h"#include "stdlib.h"#include "stdio.h"void main(){FILE *fp;int a[10],i,j;if((fp=fopen("d1.dat","w"))==NULL){cout<<"can't open d1.\n";exit(1);}for(i=0;i<10;i++)a[i]=rand();for(i=0;i<9;i++)for(j=0;j<9-i;j++)if(a[j]<a[j+1]){int t=a[j];a[j]=a[j+1];a[j+1]=t;}for(i=0;i<10;i++)fprintf(fp,"%d\t",a[i]);fclose(fp);}3、#include "iostream.h"#include "stdlib.h"#include "stdio.h"#include "string.h"void main(){FILE *fp;char s[100];int a[26]={0},i;if((fp=fopen("f.txt","r"))==NULL){cout<<"can't open f.\n";exit(1);}fgets(s,100,fp);strlwr(s);i=0;while(s[i]!='\0'){if(s[i]>='a'&&s[i]<='z')a[s[i]-'a']++;i++;}for(i=0;i<26;i++)cout<<(char)('a'+i)<<":"<<a[i]<<endl;fclose(fp);}4、#include "iostream.h"#include "stdlib.h"#include "stdio.h"struct student{char num[8];char name[20];double s[3],ave;}st[5];void main(){student stu[5];FILE *fp;int i;if((fp=fopen("stu","wb"))==NULL){cout<<"can't open stu.\n";exit(1);}for(i=0;i<5;i++){cin>>stu[i].num>>stu[i].name>>stu[i].s[0]>>stu[i].s[1]>>stu[i].s[2];stu[i].ave=(stu[i].s[0]+stu[i].s[1]+stu[i].s[2])/3;}fwrite(stu,sizeof(student),5,fp);fclose(fp);}第八章一、选择题1、A2、A3、C4、C5、6、D7、B8、D9、C 10、B11、D 12、A 13、B 14、D 15、D二、阅读程序写结果1、con1 calledcon2 calledcon3 calleda=0,b=0a=10,b=10a=10,b=202、0 51 52 53 54 53、4564、10,106,67,95、x=0x=10x=7三、编程题1、#include "iostream.h"#include "string.h"class Cat{private:int age;double weight;char color[10];public:void set(int a,double w,char c[10]){age=a;weight=w;strcpy(color,c);}void print(){cout<<"age:"<<age<<"\tweight:"<<weight<<"\color:"<<color<<endl;}int getage(){return age;}double getweight(){return weight;}char *getcolor(){return color;}};void main(){Cat c1;int a;double w;char c[10];cin>>a>>w>>c;c1.set(a,w,c);c1.print();}2、#include "iostream.h"#include "string.h"#include "stdio.h"class Mystring{private:char *str;public:Mystring(){ }Mystring(char *s){str=new char[100];strcpy(str,s);}void set(char *s){str=new char[100];strcpy(str,s);}void print(){cout<<str<<endl;}int length( ){int i=0;while(str[i]!='\0')i++;return i;}char *stringcat(Mystring s2){int i=0,j=0;while(str[i]!='\0')i++;while(s2.str[j]!='\0')str[i++]=s2.str[j++];str[i]='\0';return str;}};void main(){Mystring s1;char s[100];gets(s);s1.set(s);s1.print();gets(s);Mystring s2(s);s2.print();cout<<s1.stringcat(s2);}3、#include "iostream.h"#include "string.h"#include "stdio.h"class Point{private:double x,y;public:Point(){ }Point(double x1,double y1){x=x1;y=y1;}void set(double x1,double y1){x=x1;y=y1;}void move(double x1,double y1){x=x+x1;y=y+y1;}void print(){cout<<"("<<x<<","<<y<<")"<<endl;}double getx(){ return x;}double gety(){ return y;}};class Circle:public Point{private:double r;public:void set(double x1,double y1,double r1){Point::set(x1,y1);r=r1;}void print(){Point::print();cout<<r<<endl;}double getr(){ return r;}double s(){ return 3.14*r*r;}};void main(){Point p1;Circle c1;double x1,y1,r1;cin>>x1>>y1;p1.set (x1,y1);p1.print ();cin>>x1>>y1;p1.move (x1,y1);p1.print ();cin>>x1>>y1>>r1;c1.set (x1,y1,r1);c1.print ();cout<<c1.s()<<endl;}4、#include "iostream.h"class Point{private:int x,y;public:Point(){ }Point(int x1,int y1){ x=x1; y=y1;}friend Point operator+(Point &p1,Point &p2){Point p(p1.x+p2.x,p1.y+p2.y);return p;}void print(){cout<<"("<<x<<","<<y<<")"<<endl;}};void main(){Point p1(1,2),p2(2,3),p3;p3=p1+p2;p3.print();}5、#include "iostream.h"#include "stdio.h"#include "string.h"class Teacher{private:char name[20];public:void virtual input(){cout<<"input name:";cin>>name;}virtual double wage()=0;void virtual print(){cout<<name<<'\t';}};class Professor:public Teacher{private:int cnum; //课时数public:void input(){Teacher::input ();cout<<"input cnum:";cin>>cnum;}double wage(){return (3000+40*cnum);}void print(){Teacher::print ();cout<<wage()<<endl;}};class ViceProfessor:public Teacher {private:int cnum; //课时数public:void input(){Teacher::input ();cout<<"input cnum:";cin>>cnum;}double wage(){return (2500+30*cnum);}void print(){Teacher::print ();cout<<wage()<<endl;}};class Lecture:public Teacher{private:int cnum; //课时数public:void input(){Teacher::input ();cout<<"input cnum:";cin>>cnum;}double wage(){return (2000+25*cnum);}void print(){Teacher::print ();cout<<wage()<<endl;}};void main(){Teacher *t;Professor *p;ViceProfessor *vp;Lecture *l;char ptitle[20],ch;//职称do{cout<<"input ptitle(professor,viceprofessor or lecture):"<<endl;gets(ptitle);if(!strcmp(ptitle,"professor")){p=new Professor;t=p;}else if(!strcmp(ptitle,"viceprofessor")){vp=new ViceProfessor;t=vp;}else{l=new Lecture;t=l;}t->input ();t->print ();cout<<"continue(y/n)?";cin>>ch;}while(ch=='y');}。

《C语言程序设计（Visual C++6.0环境）》习题答案习题六一、思考题1、编写程序，将10个整形数2、4、6，…18，20赋给一个数组，然后使用指针输出显示该数组各元素的值。

#include “stdio.h”main(){int i,*p;int a[10]={2,4,6,8,10,12,14,16,18,20};p=a;for(i=0;i<10;i++)printf(“%3d”,*(p+i));}2、编写程序，将两个字符串连接起来。

#include<stdio.h>#define N 120main(){ char s1[N+N],s2[N],*p,*q;printf("输入2个字符串\n");scanf("%s%s",s1,s2);for(p=s1;* p!='\0'; p++);for(q=s2;*p++=*q++;);printf("两字符串连接后：%s\n",s1);}3、输入5个字符串，按英文字典排序从小到大顺序输出。

#include "stdio.h"#include "string.h"void sort(char *name[],int count){char *temp;int i,j,min;for(i=0;i<count-1;i++){min=i;for(j=i+1;j<count;j++)if(strcmp(name[min],name[j])>0)min=j;if(min!=i){temp=name[i];name[i]=name[min];name[min]=temp;}}}main(){char *name[5]={"BASIC","FORTRON","PASAL","C","FOXBASE"};int i=0;sort(name,5);for(;i<5;i++)printf("%s\n",name[i]);}4、编一程序，输入月份号，输出该月的英文月名。

C语言程序设计第四版第六章答案_谭浩强1、用筛选法求100之内的素数。

解：#include#includeint 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++)< bdsfid="73" p=""></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("\");for (i=2,n=0;i<=100;i++){ if(a[i]!=0){printf("%5d",a[i]);n++;}if(n==10){printf("\");n=0;}}printf("\");return 0;}2、用选择法对10整数排序。

解：#includeint main(){int i,j,min,temp,a[11];printf("enter data:\");for (i=1;i<=10;i++){printf("a[%d]=",i);scanf("%d",&a[i]);}printf("\");printf("The orginal numbers:\");for (i=1;i<=10;i++)printf("%5d",a[i]);printf("\");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("\The sorted numbers:\");for (i=1;i<=10;i++)printf("%5d",a[i]);printf("\");return 0;}3、求一个3×3的整型矩阵对角线元素之和。

C程序设计（第五版）-第6章利⽤数组处理批量数据课后习题答案1.⽤筛选法求100质数⼜称素数。

⼀个⼤于1的⾃然数，除了1和它⾃⾝外，不能被其他⾃然数整除的数叫做质数；否则称为合数（规定1既不是质数也不是合数）先解释⼀下筛选法的步骤：<1> 先将1挖掉(因为1不是素数)。

<2> ⽤2去除它后⾯的各个数，把能被2整除的数挖掉，即把2的倍数挖掉。

<3> ⽤3去除它后⾯的各数，把3的倍数挖掉。

<4> 分别⽤5…各数作为除数去除这些数以后的各数。

上述操作需要⼀个很⼤的容器去装载所有数的集合，只要满⾜上述条件，即2的倍数⼤于1的全部置0，3的倍数⼤于1的全部置0，4的倍数⼤于1的全部置0……⼀直到这个数据集合的末尾，这样⼀来不为0的数就是素数了，然后按下标在⾥⾯进⾏查找就好了1#include <stdio.h>2#include <windows.h>3int main()4{5printf("------------------\n");6int i, j, k, a[100];7// 先给100个数赋值8for (i = 0; i < 100; i++)9{10a[i] = i + 1;11}1213// 1不是质数也不是合数14a[0] = 0;1516for (i = 0; i < 100; i++)17{18for (j = i + 1; j < 100; j++)19{20// 把后⾯的数能整除前⾯的数赋值为021if (a[i] != 0 && a[j] != 0)22{23if (a[j] % a[i] == 0)24{25a[j] = 0; //把不是素数的都赋值为026}27}28}29}3031// 打印质数，每10个换⾏32for (i = 0; i < 100; i++)33{34if (k % 10 == 0)35{36printf("\n");37}38if (a[i] != 0)39{40printf("%d ", a[i]);41k++;42}43}4445return 0;46}2.⽤选择法对101#include <stdio.h>2#include <windows.h>3int main()4{5printf("请输⼊10个数：\n");6int minIndex, temp, a[10];78for (int i = 0; i < 10; i++)9{10scanf("%d", &a[i]);11}1213for (int i = 0; i < 10; i++)14{15minIndex = i;16for (int j = i + 1; j < 10; j++)17{18if (a[j] <= a[minIndex])19{20minIndex = j;21}22}2324temp = a[i];25a[i] = a[minIndex];26a[minIndex] = temp;27}2829printf("排序后结果：\n");3031for (int i = 0; i < 10; i++)32{33printf("%d ", a[i]);34}35return 0;36}3.求⼀个3*31#include <stdio.h>2#include <windows.h>3int main()4{5printf("请输⼊元素：\n");6int x, y, z, a[3][3];7for (int i = 0; i < 3; i++)8{9for (int j = 0; j < 3; j++)10{11scanf("%d", &a[i][j]);12}13}14printf("输出刚刚输⼊的元素：\n");15for (int i = 0; i <= 2; i++)16{17for (int j = 0; j <= 2; j++)18{19printf("%d\t", a[i][j]);20}2122printf("\n");23}24printf("\n");25// 计算对⾓线的合26for (int i = 0; i < 3; i++)27{28x += a[i][i];29}3031for (int i = 0, j = 2; i < 3; i++, j--)32{33y += a[i][j];34}35z = x + y;36printf("左上到右下对⾓线的合：%d\n", x); 37printf("右上到左下对⾓线的合：%d\n", y); 38printf("两条对⾓线之合：%d\n", z);39// 结果40// 请输⼊元素：41// 1 2 3 4 5 6 7 8 942// 输出刚刚输⼊的元素：43// 1 2 344// 4 5 645// 7 8 94647// 左上到右下对⾓线的合：1548// 右上到左下对⾓线的合：3149// 两条对⾓线之合：4650return 0;51}4.1#include <stdio.h>2#include <windows.h>3int main()4{5printf("------------------\n");6int t, x, a[5] = {1, 2, 4, 5, 6};78printf("请输⼊需要插⼊的数字：\n");9scanf("%d", &x);10for (int i = 0; i < 5; i++)11{12if (x < a[i])13{14t = a[i];15a[i] = x;16x = t;17}18printf("%3d", a[i]);19}20printf("%3d", x);2122return 0;23}5.讲⼀个数组的值按逆序重新存放。

C语⾔程序设计教程第六章课后习题参考答案P158 1求三个实数最⼤值#includefloat max(float,float,float);int main(){float a,b,c,m;printf("请输⼊三个实数:");scanf("%f %f %f",&a,&b,&c);printf("最⼤数为%f\n",max(a,b,c));return 0;}float max(float a,float b,float c){float t;if(a>b&&a>c)t=a;else if(b>a&&b>c)t=b;elset=c;return t;}P158 2求最⼤公约数最⼩公倍数#includeint fun1(int a,int b);int fun2(int a,int b);int main(){int a,b;printf("请输⼊两个整数:");scanf("%d %d",&a,&b);printf("最⼤公约数为:%d\n",fun1(a,b));int t,r;if(a{t=a;a=b;b=t;}while((r=(a%b))!=0) {a=b;b=r;}return b;}int fun2(int a,int b) {int n;n=(a*b)/fun1(a,b); return n;}P158 3求完全数#includevoid wan(int n); void main(){int n;for(n=1;n<1000;n++) wan(n);printf("\n");}void wan(int n){if(n%i==0)s=s+i;}if(n==s)printf("%d\t",n); }P158 4⽆暇素数#include#includeint nixvshu(int n);int isPrime(int n);int main(){int n,t;printf("⽆暇素数:\n");for(n=100;n<=999;n++) {t=nixvshu(n);if(isPrime(n)&&isPrime(t)) printf("%d\t",n);}printf("\n");return 0;}int nixvshu(int n){int x=0;while(n){x=x*10+n%10;n=n/10;}return x;int i;for(i=2;i<=sqrt(n);i++)if(n%i==0) return 0;return n;}P158 7递归函数#includeint Y(int n){if(n==0)return 0;if(n==1)return 1;if(n==2)return 2;elsereturn (Y(n-1)+Y(n-2)+Y(n-3)); } int main(){int n,k=0;for(n=0;n<=19;n++){printf("%d\t",Y(n));k++;if(k%5==0)printf("\n");}return 0;}P124 6分解质因数#include#includevoid fenjie(int );int main(){int n;printf("请输⼊⼀个正整数:"); scanf("%d",&n);if(isPrime(n)){printf("%d=1*%d\n",n,n);}else{fenjie(n);printf("\n");}return 0;}int isPrime(int n){int i;for(i=2;i<=sqrt(n);i++){if(n%i==0) return 0;}return 1;}int zhi(int n){int m;for(m=2;m<=n;m++){if(isPrime(m)&&(n%m==0)) break;void fenjie(int n){int m;printf("%d=%d",n,zhi(n));while(n>zhi(n)){m=zhi(n);n=n/m;printf("*%d",zhi(n));}}P47 1输出闰年#includeint f(int year);int main(){int year,k=0;for(year=1900;year<=2000;year++){if(f(year)){printf("%d\t",year);k++;if(k%5==0)printf("\n");}}return 0;}int f(int year){if((year%4==0&&year%100!=0)||(year%400==0)) return year;P47 2输出回⽂数#includeint fun(int n);int main(){int n,k=0;for(n=10;n<=2000;n++) {if(n == fun(n)){printf("%d\t",n);k++;if(k%5==0)printf("\n");}}return 0;}int fun(int n){int i=0;while(n>0){i=i*10+n%10;n=n/10;}return i;}P47 3进制转换#includevoid trans(int n,int base); int main()printf("请输⼊要转换的⼗进制数："); scanf("%d",&n);printf("请输⼊转换的进制："); scanf("%d",&base);trans(n,base);printf("\n");return 0;}（⽅法⼀）void trans(int n,int base){if(n){trans(n/base,base);if(n%base>=10)switch(n%base){case 10:printf("A");break;case 11:printf("B");break;case 12:printf("C");break;case 13:printf("D");break;case 14:printf("E");break;case 15:printf("F");break;}elseprintf("%d",n%base);}}（⽅法⼆）void trans(int n,int base){int r;char c;trans(n/base,base); r=n%base;if(r>=10)c='A'+r-10;elsec='0'+r;printf("%c",c);}}。

# include<stdio.h># include<math.h># include<string.h>/*int main() //筛选法求100以内的素数{intp,m=0,i,j;for(i=2;i<100;i++){p=(int)sqrt(i);for(j=2;j<=p;j++)if(i%j==0)break;if(j==p+1){printf("%-3d",i);m+=1;}if(m%10==0)printf("\n");}putchar('\n');}void sort(int a[],int n) //选择法排序{inti,j,k;for(i=1;i<n;i++)for(j=i;j<n;j++)if(a[i-1]>a[j]){k=a[i-1];a[i-1]=a[j];a[j]=k;}}int main(){int a[10],i;printf("10 munber:");for(i=0;i<10;i++)scanf("%d",&a[i]);sort(a,10);printf("sort:");for(i=0;i<10;i++)printf("%-2d",a[i]);putchar('\n');}int main(){int a[3][3],i,j,s=0; //求对角线元素之和printf("输入矩阵:\n");for(i=0;i<3;i++)for(j=0;j<3;j++)scanf("%d",&a[i][j]);for(i=0;i<3;i++)s+=a[i][i];printf("对角线元素之和为:%d\n",s);}int main(){int a[]={1,2,3,4,6,7,8,9}; //向已排好序的数组插入新数ints,b,i,j;s=sizeof(a)/4;printf("enter number:");scanf("%d",&b);for(i=0;i<s;i++)if(b>a[i]&&i!=s-1)continue;else if(i==s-1)a[s]=b;else{for(j=s;j>i;j--)a[j]=a[j-1];a[i]=b;break;}printf("new sort:");for(i=0;i<=s;i++)printf("%-2d",a[i]);printf("\n");}int main(){int a[9]={1,2,3,4,5,6,7,8,9},i,n,t,l; //逆序输出l=sizeof(a)/4;n=l/2;for(i=0;i<=n;i++){t=a[i];a[i]=a[l-i-1];a[l-i-1]=t;}for(i=0;i<l;i++)printf("%d ",a[i]);putchar('\n');}int main(){int a[10][10],i,j; //杨辉三角for(i=0;i<10;i++)for(j=0;j<10;j++)if(j==0||i==j)a[i][j]=1;else if(i>j)a[i][j]=a[i-1][j-1]+a[i-1][j];for(i=0;i<10;i++)for(j=0;j<10;j++){if(i>=j)printf("%-4d",a[i][j]);if(j==9)putchar('\n');}putchar('\n');}int a[25][25]={0}; //魔方阵void array(int n){staticint b[2];inti,j,k;for(k=1;k<=n*n;k++){if(k==1){a[0][n/2]=k;b[0]=0;b[1]=n/2;}else{if(b[0]==0&&b[1]<n-1){a[n-1][(b[1]+1)]=k;b[0]=n-1;b[1]=b[1]+1;}else if(b[0]>0&&b[1]==n-1){a[b[0]-1][0]=k;b[0]=b[0]-1;b[1]=0;}else if(b[0]==0&&b[1]==n-1){a[1][n-1]=k;b[0]=1;b[1]=n-1;}else if(a[b[0]-1][b[1]+1]!=0){a[b[0]+1][b[1]]=k;b[0]=b[0]+1;}else{a[(b[0]-1)][(b[1]+1)]=k;b[0]=b[0]-1;b[1]=b[1]+1;}}}for(i=0;i<n;i++)for(j=0;j<n;j++){printf("%-5d",a[i][j]);if(j==n-1)putchar('\n');}}int main(){int n;printf("输入小于25的奇数:");scanf("%d",&n);array(n);}int main()int a[3][3],i,j,k,l,max; //寻找鞍点printf("输入矩阵：\n");for(i=0;i<3;i++)for(j=0;j<3;j++)scanf("%d",&a[i][j]);for(i=0;i<3;i++){max=a[i][0];l=0;for(j=1;j<3;j++){k=0;if(max<a[i][j]){max=a[i][j];l=j;}if(j==2){for(;k<3;k++)if(max<=a[k][l])continue;elsebreak;}if(k==3)printf("鞍点为：%d\n",max);}}}void star(){printf("* * * * *\n"); //打印图案}int main(){inti,j;for(i=0;i<5;i++){for(j=0;j<2*i;j++)printf(" ");star();}#define N 15voidnum(int a[],int n){int min=0,max=N-1,mid=N/2; //折半查找法while(min<=max){mid=(max+min)/2;if(n>a[mid]){min=mid+1;}else if(n<a[mid]){max=mid-1;}else{printf("第%d个数\n",mid+1);break;} }if(min>max)printf("无此数！\n");}int main(){int a[]={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15},n,i;for(i=0;i<N;i++)printf("%-3d",a[i]);printf("\nenter number:");scanf("%d",&n);num(a,n);}int main(){char a[3][80],u=0,l=0,n=0,b=0,o=0,i,j;for(i=0;i<3;i++)for(j=0;j<80;j++)scanf("%c",&a[i][j]);for(i=0;i<3;i++)for(j=0;j<80;j++){if(a[i][j]>='A'&&a[i][j]<='Z')u+=1;else if(a[i][j]>='a'&&a[i][j]<='z')l+=1;else if(a[i][j]>='0'&&a[i][j]<='9')n+=1;else if(a[i][j]==' ')b+=1;elseo+=1;}printf("\n大写字母个数：%d\n小写字母个数：%d\n数字个数：%d\n空格个数：%d\n 其他字符个数：%d\n",u,l,n,b,o);}# define N 20int main(){char a[N]="512/R olevblf",b[N];inti=0,j=0;printf("密码：");while(a[i]){printf("%c",a[i]);i++;}printf("\n原文：");while(a[j]){if(a[j]>='a'&&a[j]<='z')b[j]='a'+'z'-a[j];else if(a[j]>='A'&&a[j]<='Z')b[j]='A'+'Z'-a[j];elseb[j]=a[j];j++;}b[j]='\0';printf("%s\n",b);}#define N 20void str_cat(char a[],char b[]) //字符串连接{int s1=strlen(a),i=0;while(b[i]){a[s1+i]=b[i];i++;}a[s1+i]='\0';}int main(){char a[N]="I ",b[N]="love you!";str_cat(a,b);printf("%s\n",a);}#define N 20void str_cpy(char a[],char b[]) //字符串复制{inti=0;while(a[i]=b[i])i++;}int main(){char s1[N],s2[]="I love you!";str_cpy(s1,s2);printf("s2=%s\ns1=%s\n",s2,s1);}*/#define N 20intstr_cmp(char a[],char b[]) //字符串比较{inti=0,n;n=strlen(b);while(a[i]==b[i]){i++;if(i==n)break;}printf("%d",a[i]-b[i]);}int main(){char s1[N],s2[N];gets(s1);gets(s2);str_cmp(s1,s2);putchar('\n');}。

C语言程序设计-第三版-谭浩强主编第6—8章课后习题答案第六章循环语句6.1输入两个正数,求最大公约数和最小公倍数.#includevoidmain(){inta,b,num1,num2,temp;printf(\请输入两个正整数:\\n\canf(\if(num1temp=num1;num1=num2;num2=temp;}a=num1,b=num2;while(b!=0){temp=a%b;a=b;b=temp;}printf(\它们的最大公约数为:%d\\n\printf(\它们的最小公倍数为:%d\\n\}编译已通过6.2输入一行字符,分别统计出其中英文字母,空格,数字和其它字符的个数.解:#includevoidmain(){charc;intletter=0,pace=0,degit=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')letter++;eleif(c=='')pace++;eleif(c>='0'&&c<='9')digit++;eleother++;}printf(\其中:字母数=%d空格数=%d数字数=%d其它字符数=%d\\n\digit,other);}6.3求(n)=a+aa+aaa++aaa之值,其中a是一个数字，n表示a的位数。

解:voidmain(){inta,n,count=1,n=0,tn=0;printf(\请输入a和n的值:\\n\canf(\printf(\while(count<=n){tn=tn+a;n=n+tn;a=a某10;++count;}printf(\\\n\}6.4求(即1+2!+3!+4!++20!)voidmain(){floatn,=0,t=1;for(n=1;n<=20;n++){t=t某n;=+t;}printf(\\\n\}阶乘利用递归，再求和：#includeuingnamepacetd;longFunc(intn){ if(1==n)returnn;if(n>1)returnn某Func(n-1);}main(){long=0;inti=1;while(i<=6){=+Func(i);i++;}cout<6.5求voidmain(){intk,N1=100,N2=50,N3=10;float1=0.0,2=0.0,3=0.0;for(k=1;k<=N1;k++)/某计算1到100的和某/{1=1+k;}for(k=1;k<=N2;k++)/某计算1到50各数平方和某/{2=2+k某k;}for(k=1;k<=N3;k++)/某计算1到10各数倒数之和某/{3=3+1.0/k;}printf(\总和=%8.2f\\n\}已通过intmain(){intk=1,i=11,j=51;float=0.0;while(k<=10){=+k+k某k+1.0/k;while(k==10&&i<=50){=+i+i某i;while(i=50&&j<=100){ =+j;j++;}i++;}k++;}}6.6所谓“水仙开数”是指一个3位数，其个位数字立方和等于该数本身。

6-3 (1)#include <iostream>using namespace std;int main(void){int num[10], *p, i, j, temp; cout«**输入10 个整数"vvendl; for(i=0; i<10; i++){ cin»num[订;}for(i=0; i<10; ■++){// 0, 1_________ 8p = &num[i];for(j = i+1; J<10; j++){//1, 2・.・ 9 if(num[i]<=num[J]){p = &num[j];}}cout«*p«endl; temp = *p;*p = num[i]; num[i] = temp;}system(M pause f,);return 0;(2)注：N的数值可以在宏定义中更改，以下是W12的吋候:#include <iostream>using namespace std;#define N 12struct N0DE{int num;NODE *next;};int i:class List{private:NODE list[N];NODE *temp;public:List() {//设置一个结点组成的圈temp = Iist;Iist[0].num = 1 ;Iist[0].next = &list[1];//第一个结点己经完成for(i=1: i<(N-1); ■++){Iist[i].num = i+1:Iist[i].next = &Iist[i+1];}//第二个至倒数第二个已经完成Iist[N-1].num = N;Iist[N-1].next = &list[O];//最后一个结点已经完成void Next2(){temp = temp->next;temp = temp->next;}void Fun(){temp = & list[N-1];//得到链表的最后一个结点地址do{Next2();//当前是0,数到2〃现在temp已经是标号为2的结点的地址了cout«(temp->next)->num«ff M;//$ny出要被删除的结点(p+1) temp->next = (temp->next)->next ;//^ 被删除的结点的前一个(p)和后一个相连接(p+2)}while(temp->next != (temp->next)->next); cout«temp->num;}};int main(void){List one; one.Fun(j; system(ft pause f,); return 0;}(3)#include <iostream>#include <string> using namespace std;#define SIZE 512class Str{private:char str1[SIZE]:char str2[SIZE];char *p;int i, length;public:void set(){cout«ff输入第一个字符串:ff«endl; cin>>str1;cout«ff输人第二个字符串:ft«endl; cin>>str2;}void fun(){Iian(str1, str2):}void Iian(char *m, char *n){ length = strlen(n); for(i=0; i<strlen(n)+1; i++){ P = & n[i];str1 [length+i] = *p;}cout«ft连接之后的结果：\n ft«str1«endl;}}；int main(void){ Str one; one.set(); one.funO; system(ft pause f,); return 0;}(4)#include <iostream>#include <string>using namespace std;#define SIZE 512class Str{private:char str[SIZE]:int upper, lower, space, number, other, i:public:Str(){upper=Iower=space=number=other=0;void set(){cout«ff输入一个字符ff«endl;gets(str);}void check(){for(i=0; i<strlen(str); i++){if(str[i]>=w A・ && str[i]<="Z>){upper++;}else if(str[i]>=w a w && str[i]<=w z w){lower++;}else if(str[i]=="・){space++;}else if(str[i]>=w O w && str[i]<=w9w){number++;}else{other++;}}}void te11(){check();coutvv"箕冇大写字母ff«upper«ft个，小写字母"«lower«ff个，空格"vvspacevv”个，数字ff«numbervv”个，其他字符■•vvothervv”个”vvendl;}};int main(void){Str one;one.set();one.tel 1();system(ft pause ff);return 0;}(5)代码中要用到strlen函数，所以包含了string头文件，但strcmp函数在string中己经定义，所以下题屮改用Strcmpo#include <iostream>#include <string>using namespace std;#define SIZE 512class Str{private:char str1[SIZE];char str2[SIZE];■ nt i;public:void get(){cout«ft输入第一个字符串vvendl;OP+Q(Q+r1\■cout«M输入菊二个字符串：“vvendl; gets(str2);}void fun(){ Strcmp(str1, str2);}void Strcmp(char *m, char *n){int num;for(i=0; i<strlen(str1) || i<strlen(str2); i++){if(str1[i]==str2[i]){num = 0;}else{num = str1[i] - str2[i]; break;}}}};irrt main(void){Str one;one.get();one・fun();system(fl pause f,); return 0;}(6)注意：原有数组是不变的，但指针数组是排序Z后的。

1、选择题（1）A（2）C（3）A（4）B（5）B（6）D（7）D（8）B（9）D（10）B2、填空题（1）a=10,b=20a=20,b=10（2）**pp=603、程序设计题（1）#include<stdio.h>char *month_name(int n);void main(){int n;printf("\nPlease enter 1 integer:");scanf("%d",&n);printf("%d month :%s\n",n,month_name(n));}char *month_name(int n){static char*name[]={"illegal month","Jan","Feb","Mar","Apr","May","Jun","July","Aug","Sept","Oct","Nov","Dec"};return ((n<1||n>12)?name[0]:name[n]);}（2）#include<stdio.h>#define N 10sort(int data[]){int i,j,min_a,temp;for(i=0;i<N;i++){min_a=i;for(j=i+1;j<N;j++)if(*(data+j)<*(data+min_a))min_a=j;if(min_a!=i){temp=*(data+min_a);*(data+min_a)=*(data+i);*(data+i)=temp;}}}main(){int i,j,data[N],temp;int min_a;printf("\nPlease input %d int:\n",N);for(i=0;i<N;i++)scanf("%d",&data[i]);sort(data);printf("After sorted:\n");for(i=0;i<N;i++)printf(" %d",data[i]);}（3）#include <stdlib.h>void reverse(char *c);void main(){char str[80];puts("Please enter 1 string\n");gets(str);reverse(str) ;puts("After reversed\n");puts(str);}void reverse(char *c){char *p,*q,temp;int size=0;for(p=c;*p!='\0';p++)size++;size=size/2;for(q=c,p--;q<c+size;q++,p--){temp=*q;*q=*p;*p=temp;}}（4）#include<stdio.h>#include<string.h>void sort(char *keyword[],int size);void print(char *keyword[],int size)void main(){char *keyword[]={"if","else","case","switch","do","whlie","for","break","continue"};sort(keyword,9);print(keyword,9);}void sort(char *keyword[],int size){int i,j,min_location;char *temp;for(i=0;i<size-1;i++){min_location=i;for(j=i+1;j<size;j++)if(strcmp(keyword[min_location],keyword[j])>0) min_location=j;if(min_location!=i){temp=keyword[i];keyword[i]=keyword[min_location];keyword[min_location]=temp;}}}void print(char *keyword[],int size){int i;for(i=0;i<size;i++)printf("\n%s",*(keyword+i));}（5）#include<stdio.h>void fun_char(char str1[],char str2[],char str3[]);void main(){char str1[80],str2[80],str3[80],c,i;printf("\nPlease enter 2 string:");scanf("%s%s",str1,str2);fun_char(str1,str2,str3);printf("Third string is %s.",str3);}void fun_char(char *str1,char *str2,char *str3){int i,j,k,flag;i=0,k=0;while(*(str1+i)!='\0'){j=0;flag=1;while(*(str2+j)!='\0'&&flag==1){if(*(str2+j)==*(str1+i)) flag=0;j++;}if(flag){*(str3+k)=*(str1+i); k++;}i++;}*(str3+k)='\0';}（6）#include<stdio.h>int count_word(char *str);void main(){char str1[80],c,res;puts("\nPlease enter a string:");gets(str1);printf("There are %d words in this sentence",count_word(str1)); }int count_word(char *str){int count ,flag;char *p;count=0;flag=0;p=str;while(*p!='\0'){if(*p==' ')flag=0;else if(flag==0){flag=1;count++;}p++;}return count;}（7）#include<stdio.h>#include<string.h>char *encrypt(char *string);char *decrypt(char *string);main(){char item[80];char *point;char *pEncrypted;char *pDecrype;printf("Please enter the string need to encrypt:\n");gets(item);point=item;pEncrypted=encrypt(point);printf("\nThe string after encrypted is:\n%s\n",pEncrypted); pDecrype=decrypt(pEncrypted);printf("\nThe string after decrypted is:\n%s\n",pDecrype);free(pEncrypted);free(pDecrype);}char *encrypt(char *string){char *q,*t;q=(char *)malloc(sizeof(char)*80);if(!q){printf("No place to malloc!");return 0;}t=q;while(*string!='\0'){*q=*string-2;string++;q++;}*q='\0';return t;}char *decrypt(char *string){char *q,*t;q=(char *)malloc(sizeof(char)*80); if(!q){printf("No place to malloc!");return 0;}t=q;while(*string!='\0'){*q=*string+2;string++;q++;}*q='\0';return t;}。

c语⾔程序设计第五版课后答案谭浩强第六章习题答案第六章：利⽤数组处理批量数据1. ⽤筛选法求100之内的素数【答案解析】素数：约数为1和该数本⾝的数字称为素数，即质数筛选法：⼜称为筛法。

先把N个⾃然数按次序排列起来。

1不是质数，也不是合数，要划去。

第⼆个数2是质数留下来，⽽把2后⾯所有能被2整除的数都划去。

2后⾯第⼀个没划去的数是3，把3留下，再把3后⾯所有能被3整除的数都划去。

3后⾯第⼀个没划去的数是5，把5留下，再把5后⾯所有能被5整除的数都划去。

这样⼀直做下去，就会把不超过N 的全部合数都筛掉，留下的就是不超过N的全部质数。

因为希腊⼈是把数写在涂腊的板上，每要划去⼀个数，就在上⾯记以⼩点，寻求质数的⼯作完毕后，这许多⼩点就像⼀个筛⼦，所以就把埃拉托斯特尼的⽅法叫做“埃拉托斯特尼筛”，简称“筛法”。

（另⼀种解释是当时的数写在纸草上，每要划去⼀个数，就把这个数挖去，寻求质数的⼯作完毕后，这许多⼩洞就像⼀个筛⼦。

）【代码实现】//⽤筛选法求100以内的素数#include<stdio.h>int main(){int i, j, k = 0;// 将数组汇总每个元素设置为：1~100int a[100];for (i = 0; i < 100; i++)a[i] = i+1;// 因为1不是素数，把a[0]⽤0标记// 最后⼀个位置数字是100，100不是素数，因此循环可以少循环⼀次a[0] = 0;for (i = 0; i < 99; i++){// ⽤a[i]位置的数字去模i位置之后的所有数据// 如果能够整除则⼀定不是素数，该位置数据⽤0填充for (j = i + 1; j < 100; j++){if (a[i] != 0 && a[j] != 0){//把不是素数的都赋值为0if (a[j] % a[i] == 0)a[j] = 0;}}}printf(" 筛选法求出100以内的素数为：\n");for (i = 0; i < 100; i++){//数组中不为0的数即为素数if (a[i] != 0)printf("%3d", a[i]);}printf("\n");return 0;}【运⾏结果】2. ⽤选择法对10个整数排序【答案解析】选择排序原理：总共两个循环，外循环控制选择的趟数，内循环控制具体选择的⽅式。

C++程序设计基础课后答案第六章6.1 阅读下列程序，写出执行结果1. #includeclass T{ public :T() { a = 0; b = 0; c = 0; }T( int i , int j , int k ){ a = i; b =j ; c = k; }void get( int &i , int &j , int &k ) { i = a; j = b; k = c; }T operator * ( T obj );private:int a , b , c;};T T::operator * ( T obj ){ T tempobj;tempobj.a = a * obj.a;tempobj.b = b * obj.b;tempobj.c = c * obj.c;return tempobj;}void main(){ T obj1( 1,2,3 ), obj2( 5,5,5 ), obj3; int a , b , c;obj3 = obj1 * obj2;obj3.get( a, b, c );cout << "( obj1 * obj2 ): \t"<< "a = " << a << '\t' << "b = " << b<< '\t' << "c = " << c << '\t' << endl; ( obj2 * obj3 ).get( a, b, c );cout << "( obj2 * obj3 ): \t "<< "a = " << a << '\t' << "b = " << b << '\t' << "c = "<< c<< '\t' << endl; }2. #include < iostream.h >class Vector{ public:Vector(){ }Vector(int i,int j){ x = i ; y = j ; }friend Vector operator + ( Vector v1, Vector v2 ){ Vector tempVector ;tempVector.x = v1.x + v2.x ;tempVector.y = v1.y + v2.y ;return tempVector ;}void display(){ cout << "( " << x << ", " << y << ") "<< endl ; } private:int x , y ;};void main(){ Vector v1( 1, 2 ), v2( 3, 4 ), v3 ;cout << "v1 = " ;v1.display() ;cout << "v2 = " ;v2.display() ;v3 = v1 + v2 ;cout << "v3 = v1 + v2 = " ;v3.display() ;}6.2 思考题1．一个运算符重载函数被定义为成员函数或友员函数，从定义方式、解释方式和调用方式上有何区别？可能会出现什么问题？请用一个实例说明之。