实验教程编程答案-副本

1、实验二 输入/输出与顺序结构4. 编程题 输入学生的语文、数学、英语、物理4门课程的成绩，计算该学生的总成绩和平均成绩并输出。#include <iostream.h>void main() double eng, chin,math,phy,sum,aver;cout<<"please input 4 scores: "cin>>eng>>chin>>math>>phy; /输入成绩sum=eng+chin+math+phy; /计算总成绩aver=sum/4; /计算平均分cout<<&

2、quot;Sum="<<sum<<endl<<"tAverage="<<aver<<endl; /输出 编写程序，从键盘输入一个大写英文字母，输出对应的小写字母。#include <iostream.h>void main()char c1,c2;cout<<"Please input an upper letter: "cin>>c1;c2=c1+32;cout<<"c1="<<c1<<&qu

3、ot;tc2="<<c2<<endl;实验三 选择结构程序设计3. 编程题XX 由键盘输入三个字符，输出其中的最大者。XX【源程序】#include <iostream.h>void main() char x,max; cout<<"Please input three characterss: " cin>>x; max=x; cin>>x; if (x>max) max=x; cin>>x; if (x>max) max=x; cout<<"

4、max = "<<max<<endl; 编程求下面符号函数值：y =0 （x=0）1 （x>0）-1 （x<0）【源程序】#include <iostream.h>void main()int x,y;cout<<"Please input x= "cin>>x;if (x>0)y=1;else if (x=0)y=0;elsey=-1;cout<<"y="<<y<<endl; 计算奖金。设企业利润为L，当企业利润L不超过5000元

5、时，奖金为利润的1.5%，当5000L10000元时，超过5000元部分奖金为2%（5000元以下仍按1.5%）；当10000L20000元，除10000以下的按上述方法计算外，超过10000元部分按2.5%计算奖金；如果20000L50000元，超过20000元部分按3%计算奖金；当50000L100000元时，超过50000元部分按3.5%计算奖金；当L超过100000元时，超过100000元部分按4%计算奖金。由键盘输入L的值，编程计算相应的奖金并输出。【源程序】#include<iostream.h>void main()double L,S;cout<<&qu

6、ot;please input L="cin>>L;if(L<5000) S=L*0.015;else if(L<10000)S=75+(L-5000)*0.02; else if(L<20000)S=175+(L-10000)*0.025;else if(L<50000)S=175+250+(L-20000)*0.03;else if(L<100000)S=175+250+900+(L-50000)*0.035;elseS=175+250+900+1750+(L-100000)*0.04;cout<<"S="

7、<<S<<endl; 输入年龄，输出所处人群：9岁以下为儿童，输出A；1019为少年，输出B；2029为青年，输出C；3049为中年，输出D；50以上为老年，输出E。【源程序】#include <iostream.h>void main()int age;cout<<"Please input age: "cin>>age;switch(age/10)case 0:cout<<"A-儿童n"break;case 1:cout<<"B-少年n"break

8、;case 2:cout<<"C-青年n"break;case 3:case 4:cout<<"D-中年n"break;default:cout<<"E-老年n"break;实验四 循环结构程序设计4编程题 输入n，求1+2+3+n的和。 #include<iostream.h>void main() int i,n;double sum=0;cin>>n;for(i=1;i<=n;i+) sum+=i;cout<<"1+2+3+.+"&

9、lt;<n<<"="<<sum<<endl;实验五 典型结构程序设计2编程（1）输入10个字符，输出其中的最大者。#include<iostream.h>void main()char ch,maxchar;cout<<"please input ten character:"cin>>ch;maxchar=ch;for(int i=1;i<10;i+) cin>>ch; if(ch>maxchar)maxchar=ch;cout<<&quo

10、t;maxchar="<<maxchar<<endl;（2）一个球从100m高度自由落下，每次落地后反弹回原来高度的一半，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？分析：共经过： 100*(1+1/2+1/4+1/8-+1/1024) 米第10次：100/1024米#include<iostream.h>void main()double s=1,t=1,sum,t10;int i;for(i=1;i<=10;i+)t=2*t;s=s+1/t;t10=100/t;sum=100*s;cout<<&quo

11、t;sum="<<sum<<"tt10="<<t10<<endl;（4）编写程序，对输入的一批整数统计出正数的个数、负数的个数、奇数的个数、偶数的个数，要求所统计的整数由键盘输入，以0作为输入数据结束的标志。#include<iostream.h>void main()int a=0,b=0,c=0,d=0,x;cin>>x;while(x!=0)if(x>0)a+=1;if(x<0)b+=1;if(x%2)c+=1;else d+=1;cin>>x;cout<&

12、lt;"正数个数="<<a<<endl;cout<<"负数个数="<<b<<endl;cout<<"奇数个数="<<c<<endl;cout<<"偶数个数="<<d<<endl;实验六 * 输入10个学生的成绩，求其平均值，输出最高成绩，并统计低于平均值的人数。 * 注： LT是小于的意思，less than */#include <iostream.h>const dou

13、ble LOWER = - 10000;void main()double score10;double highestScore = LOWER;double average = 0;double numLTaverage = 0;/输入、求最高分、求总分for (int i = 0; i < 10 ; + i)cout << "Please input the score ( " << i + 1 << "/10):"cin >> scorei;if ( highestScore < sco

14、rei ) highestScore = scorei;average += scorei;average /= 10;/求成绩低于平均分的人数for ( i = 0; i < 10; + i)if ( scorei < average ) + numLTaverage ;cout << "The average score is " << average << endl;cout << "The highest score is " << highestScore <<

15、 endl;cout << "The number of LT average is " << numLTaverage << endl;/* * 分别用冒泡法和选择法对输入的10个整数按由大到小排序。 * 冒泡法参见课本第四章课后作业第四题 */#include <iostream.h>void main()int myArray10;int outer, inner;int imax;for ( int i = 0; i < 10 ; + i )cout << "Input a number:&

16、quot; ;cin >> myArrayi;for ( outer = 0 ; outer < 10 ; + outer)/在下标位outer到9之间的元素中寻找最大值imax = outer ;for ( inner = outer + 1; inner < 10 ; + inner)if ( myArrayimax < myArrayinner ) imax = inner;/将最大值与下标为outer的元素交换int temp = myArrayouter;myArrayouter = myArrayimax;myArrayimax = temp;/输出f

17、or ( i = 0; i < 10 ; + i )cout << myArrayi << ' 'cout << endl;/* * 使用折半查找法，在给定的数组中查找某个数据。 */#include <iostream.h>const int N = 10;void main()int myArrayN = 1, 5, 8, 13, 16, 34, 67, 78, 90, 100;int iSearch;cout << "Please tell me the number which you want

18、 to search: " ;cin >> iSearch;int low = 0;int high = N - 1;int mid;/开始二分查找while ( low <= high )mid = ( low + high ) / 2;if ( myArraymid = iSearch ) break;else if ( myArraymid < iSearch )low = mid + 1;elsehigh = mid - 1;/判断是否找到if ( myArraymid = iSearch )cout << "We have fo

19、und the number." << endl;elsecout << "We havn't found the number." << endl;/* * 按杨辉三角形的规律打印以下的数据（要求只打印出10行）。 * 1 * 1 1 * 1 2 1 * 1 3 3 1 * 1 4 6 4 1 * 1 5 10 10 5 1 * */#include <iostream.h>void main()int yangHui1010;yangHui00 = yangHui10 = yangHui11 = 1;fo

20、r ( int i = 2; i < 10 ; + i )yangHuii0 = yangHuiii = 1;for ( int j = 1 ; j < i ; + j )yangHuiij = yangHuii - 1j + yangHuii - 1j - 1;for ( i = 0 ; i < 10 ; + i )for (int j = 0; j <= i ; + j )cout << yangHuiij << 't'cout << endl;/*XXX 编写程序统计某班英语、语文、数学3门课程的成绩， * 学生

21、人数与成绩由键盘输入，要求统计出每门课程全班 * 的总成绩和平均成绩以及每个学生三门课程的总成绩和 * 平均成绩。 */#include <iostream.h>void main()/由用户输入学生人数int numStu;cout << "Please input the number of students: "cin >> numStu;/根据用户输入的人数建立数组，其中0-2列为三门课成绩，最后一列是总分。/最后添加一行用来存储总分。int (*p)4;p = new intnumStu + 14;/将总分置0pnumStu0

22、= pnumStu1 = pnumStu2 = 0;for ( int i = 0 ; i < numStu ; + i )/将每个人的总分置0pi3 = 0;cout << "ID is " << i + 1 << ":n"/输入英语成绩cout << "Please input the score of Eng: " ;cin >> pi0;pi3 += pi0;pnumStu0 += pi0;/输入物理成绩cout << "Please in

23、put the score of Phy: " ;cin >> pi1;pi3 += pi1;pnumStu1 += pi1;/输入英语成绩cout << "Please input the score of Mat: " ;cin >> pi2;pi3 += pi2;pnumStu2 += pi2;/输出for ( i = 0 ; i < numStu ; + i )cout << "ID " << i + 1 << ":n"cout <&

24、lt; "The sum is " << pi3 ;cout << ".tThe average is " << pi3/3.0 ;cout << endl;cout << "Eng: sum=" << pnumStu0 << ",average=" << pnumStu0/double(numStu);cout << "nPhy: sum=" << pnumStu1 <

25、< ",average=" << pnumStu1/double(numStu);cout << "nMat: sum=" << pnumStu2 << ",average=" << pnumStu2/double(numStu);delete p;/* * 编写程序求对矩阵进行转置，即将元素的行列位置交换。 */#include <iostream.h>void main()int myMatrix44;/输入for ( int i = 0; i <

26、 4; + i )for ( int j = 0; j < 4; + j)cout << '(' << i + 1 << ',' << j + 1 << "):"cin >> myMatrixij;/输出转置前的数组cout << "Before exchange:n"for ( i = 0; i < 4; + i )for ( int j = 0; j < 4; + j)cout << myMatrixij

27、<< 't'cout << endl;/转置for ( i = 0 ; i < 4 ; + i)for ( int j = 0 ; j < i ; + j)/* * 编写程序求两个矩阵的乘积，若矩阵ANM与BMK相乘， *则得到矩阵C，其行列数为N×K。注意A的列数与B的行数相同 *，才可以进行乘法操作。 */#include <iostream.h>const int N = 3;const int M = 4;const int K = 5;void main()/定义数组int aNM,bMK,cNK;/输入数组A

28、cout << "Matrix A:n"for ( int i = 0; i < N; + i)for ( int j = 0; j < M; + j)cout << '(' << i + 1 << ',' << j + 1 << ")/"cout << '(' << N << ',' << M << "):"cin >&g

29、t; aij;/输入数组Bcout << "Matrix B:n"for ( i = 0; i < M; + i)for ( int j = 0; j < K; + j)cout << '(' << i + 1 << ',' << j + 1 << ")/"cout << '(' << M << ',' << K << "):"c

30、in >> bij;/计算C，并输出cout << "Matrix C=AXB:n"for ( i = 0; i < N; + i)for ( int j = 0; j < K; + j)cij = 0;for (int k = 0 ; k < M ; + k)cij += aik * bkj;cout << cij << 't'cout << endl;int temp = myMatrixij;myMatrixij = myMatrixji;myMatrixji = temp;

31、/输出转置后的数组cout << "After exchange:n"for ( i = 0; i < 4; + i )for ( int j = 0; j < 4; + j)cout << myMatrixij << 't'cout << endl;实验七 数组与指针(一)4编写程序并上机调试运行 编写程序，输入5个字符串，输出其中最大者。要求使用二维字符数组及字符串函数。#include <iostream.h>#include<string.h>const int N=

32、3;void main()char aN20;int i,max=0;for(i=0;i<N;i+)cin.getline(ai,20);for(i=1;i<N;i+)if(strcmp(ai,amax)>0)strcpy(amax,ai);cout<<amax<<endl; 编写程序，将一个字符串中的数字字符都删除。#include <iostream.h>void main()char a20;int i=0,j=0;cout<<”Please input the characters: ”;cin.getline(a,20

33、);for(i=0;ai!='0'i+)if(ai<'0'|ai>'9')aj+=ai;aj='0' cout<<a<<endl;XXXXX 编写程序，输入一行字符，统计其中有多少个单词，单词之间用一个或多个空格分隔。#include <iostream.h>void main()char a100;int i,num=0; cout<<"Please input the characters:n"cin.getline(a,100);for(i=0;

34、ai!='0'i+)while(ai=' ')i+;num+;while(ai!=' ')i+;cout<<"The number of the words are: "<<num;实验八 数组与指针(二)4编写程序并上机调试运行 编写程序，当输入17（表示星期几）时，显示相应的星期的英文名称，输入其它整数时则显示错误信息。#include"iostream.h"const int M = 2;const int N = 3;void main()char *p7 = "M

35、onday", "Tuesday", "Wednesday", "Thirsday","Friday", "Saturday", "Sunday"char num;for(;)cout<<"Please input a number:"cin>>num;if(num < '1' | num >'7')cout<<"Error!"break;cout

36、<<pnum - 49<<endl;实验九 函数及其调用2. 编程题 编写一个判断素数的函数，在主函数中由键盘输入整数的范围，并给出在该范围内的所有素数。源程序为:int fun(int n);void main() int m1,m2;cout<<"Please input the range of integer numbers:n"cin>>m1>>m2; for(int i=m1;i<m2;i+) if(fun(i) cout<<i<<" t" int fu

37、n(int n) int i; for(i=2;i<n;i+) if(n%i=0) return 0; return n; XXXXXX 编写一个函数，根据给定的年、月、日输出该日是该年的第几天。在主函数中调用该函数并输出结果，从键盘输入年、月、日的值。#include<iostream.h>int fun(int,int,int);void main()int year,month,day;cout<<"please input year,month,day:"cin>>year>>month>>day;w

38、hile(1)if(year>0&&month>=1&&month<=12&&day>=1&&day<=31)cout<<year<<"-"<<month<<"-"<<day<<" is the "<<fun(year,month,day)<<"th day of the year!"<<endl;break;els

39、ecout<<"input error!please input again:" cin>>year>>month>>day;int fun(int year,int month,int day)int ds=day;switch(month-1)case 11:ds+=30;case 10:ds+=31;case 9:ds+=30;case 8:ds+=31;case 7:ds+=31;case 6:ds+=30;case 5:ds+=31;case 4:ds+=30;case 3:ds+=31;case 2:if(year

40、%4=0&&year%100!=0|year%400=0) ds+=29;else ds+=28;case 1:ds+=31;return ds; 编写两个函数分别求2n，n!，在主函数中调用这两个函数计算21×1!+ 22×2!+2n×n!(n<10)，并在主函数中输入n的值，输出结果。#include<iostream.h>double fun1(int n);double fun2(int n);void main()int n,i;double s=0;cout<<"please input n:&q

41、uot;cin>>n;for(i=1;i<=n;i+)s+=fun1(i)*fun2(i);cout<<"s="<<s<<endl;double fun1(int n)double m=1;for(int i=1;i<=n;i+)m*=2;return m;double fun2(int n)double m=1;for(int i=1;i<=n;i+)m*=i;return m; 编写函数求sin(x)，求sin(x)的近似公式为：在主函数中输入x的值并调用该函数，输出结果。#include<iost

42、ream.h>#include<math.h>double fun(double);void main()double x,sum;cout<<"please input x:"cin>>x;sum=fun(x);cout<<"sum= "<<sum<<endl;cout<<"sin"<<x<<"="<<sin(x)<<endl;double fun(double x)doubl

43、e s=0,t=x;for(int i=1;fabs(t)>1e-6;i+=2)s+=t;t=-t*x*x/(i+1)/(i+2);return s;实验十 函数与指针2. 编程题 编写一个程序，求方程ax2+bx+c=0的根，用三个函数分别求出当b2-4ac大于0，小于0和等于0时的根。要求从主函数输入a,b,c的值并输出结果。源程序为:方法1：#include<iostream.h>#include<math.h>void root1(double,double,double,double *,double *);double root2(double,dou

44、ble);void root3(double,double,double,double *,double *);void main()double a,b,c,d,x,x1,x2;cout<<"Please input a,b,c="cin>>a>>b>>c;if(a)d=b*b-4*a*c;if(d>0) root1(a,b,d,&x1,&x2); cout<<"x1="<<x1<<"nx2="<<x2<<

45、;endl;else if(d=0) x=root2(a,b);cout<<"x="<<x<<endl;else root3(a,b,d,&x1,&x2); cout<<"x1="<<x1<<"+"<<x2<<"i"<<endl;cout<<"x2="<<x1<<"-"<<x2<<"i

46、"<<endl;else if(b) cout<<"x="<<-c/b<<"n"else if(c)cout<<"方程无根,输入的a,b,c值有误！"<<"n"elsecout<<"方程有无穷多解。"<<"n"void root1(double a, double b, double d, double * x1, double * x2)*x1=(-b+sqrt(d)/

47、(2*a);*x2=(-b-sqrt(d)/(2*a);double root2(double a, double b)double x=-b/(2*a);return x;void root3(double a, double b, double d, double * x1, double * x2)*x1=-b/(2*a);*x2=sqrt(-d)/(2*a);方法2：#include<iostream.h>#include <math.h>void root1(double,double,double,double);void root2(double,doub

48、le,double);void root3(double,double,double,double);double x1,x2;void main()double a,b,c,t;cout<<"Please input a,b,c="cin>>a>>b>>c;if(a)t=b*b-4*a*c;if (t>0)root1(a,b,c,t);cout<<"x1="<<x1<<"nx2="<<x2<<"n"e

49、lse if(t=0)root2(a,b,c);cout<<"x1=x2="<<x1<<"n"elseroot3(a,b,c,t);cout<<"x1="<<x1<<"+"<<x2<<"i"<<"n"<<"x2="<<x1<<"-"<<x2<<"i"&

50、lt;<"n"else if(b) cout<<"x="<<-c/b<<"n"else if(c)cout<<"方程无根,输入的a,b,c值有误！"<<"n"elsecout<<"方程有无穷多解。"<<"n"void root1(double a,double b,double c,double t)t=sqrt(t);x1=(-b+t)/(2*a);x2=(-b-t

51、)/(2*a);void root2(double a,double b,double c)x1=x2=-b/(2*a);void root3(double a,double b,double c,double t)t=sqrt(-t);x1=-b/(2*a);x2=t/(2*a); 编写将一个整数n转换成字符串的函数。在主函数中调用该函数并输出结果，从键盘输入n的值。例如输入123，则输出字符串“123”。n的位数可以任意。#include<iostream.h>void convert(int n);void main() int number; cout<<&qu

52、ot;please input an integer:" cin>>number; cout<<"the output is:"<<endl; if(number<0) cout<<"-" number=-number; convert(number); cout<<endl; void convert(int n) int i; char c; if(i=n/10)!=0) convert(i); c=n%10+'0' cout<<" &q

53、uot;<<c; 编写一个函数实现对数组长度为n的整型数组a的升序排序。在主函数中调用该函数并输出结果，从键盘输入数组元素的值。#include<iostream.h>const int N=50;void input(int ,int);void sort(int ,int);void output(int ,int);void main()int aN,n;cout<<"input length of the array:"cin>>n;cout<<"input the arrayn"inp

54、ut(a,n);cout<<"the array before sortedn"output(a,n);sort(a,n);cout<<"the array after sortedn"output(a,n);void input(int x,int n) int i;for(i=0;i<n;i+)cin>>xi;void sort(int x,int n) int i,j,t;for(i=0;i<n-1;i+)for(j=0;j<n-1-i;j+)if(xj>xj+1)t=xj;xj=xj+1;xj+1=t;void output(int x,int n) int i;for(i=0;i<n;i+)cout&l