软件工程第二次作业

#include <iostream> 
#include <ctime> 
using namespace std; 
int main() 
{ 
    int a,b,c,R=0,x,z,i,y,k; 
    srand(time(NULL));  
    cin>>z; 
    if (z==1) 
    { 
        cout<<"测试开始！\n"; 
        for(i=1;i<11;i++) 
        { 
            a=rand()%100; 
            b=rand()%100; 
            x=rand()%4; 
            k=rand()%10; 
            switch(x) 
            { 
            case 0: 
                { 
                    y=a+b; 
                    cout<<"请看题:\t"; 
                    cout<<a<<"+"<<b<<"="; 
                    cin>>c; 
                    if (y==c) 
                    { 
                        R+=1; 
                        cout<<"答对了,累计答对题数为"<<R<<"\n"; 
                    } 
                    else cout<<"答错了，继续努力！\n"; 
                    break; 
                } 
            case 1: 
                { 
                    if (b>a) 
                    { 
                        z=a; 
                        a=b; 
                        b=z; 
                    } 
                    y=a-b; 
                    cout<<"请看题:\t"; 
                    cout<<a<<"-"<<b<<"="; 
                    cin>>c; 
                    if (y==c) 
                    { 
                        R+=1; 
                        cout<<"答对了,累计答对题数为"<<R<<"\n"; 
                    } 
                    else cout<<"答错了，继续努力！\n"; 
                    break; 
            case 2: 
                {  
                    if (a>10) 
                    { 
                        a=a/10; 
                    } 
                    if (b>10) 
                    { 
                        b=b/10; 
                    } 
                    y=a*b; 
                    cout<<"请看题:\t"; 
                    cout<<a<<"*"<<b<<"="; 
                    cin>>c; 
                    if (y==c) 
                    { 
                        R+=1; 
                        cout<<"答对了,累计答对题数为"<<R<<"\n"; 
                    } 
                    else cout<<"答错了，继续努力！\n"; 
                    break; 
                } 
            case 3: 
                {  
                    if (b>10) 
                    { 
                        b=b/10; 
                    } 
                    if (!(a%b==0)) 
                    { 
                        a=b*k; 
                    } 
                    y=a/b; 
                    cout<<"请看题:\t"; 
                    cout<<a<<"/"<<b<<"="; 
                    cin>>c; 
                    if (y==c) 
                    { 
                        R+=1; 
                        cout<<"答对了,累计答对题数为"<<R<<"\n"; 
                    } 
                    else cout<<"答错了，继续努力！\n";
                    break; 
                } 
                } 
            } 
        } 
    } 
    else 
    { 
        cout<<"测试已完成，谢谢您的参与\n"; 
        return 0; 
    } 
    switch(R) 
    { 
    case 10: cout<<"真是个天才！\n";break; 
    case 9: cout<<"真聪明！\n";break; 
    case 8: cout<<"还不错！\n";break; 
    case 7: cout<<"多多加油！\n";break; 
    case 6: cout<<"刚及格，还得多练习！\n";break; 
    case 5: 
    case 4: 
    case 3: 
    case 2: 
    case 1: 
    case 0: cout<<"不及格！\n";break; 
    } 
    
    return 0; 
}

总结：注意逻辑的合理性，即使可以运行也可能是死循环导致程序没法实现预期的功能。