• C++作业。


    //1.1 输入两个数A,B,算出结果
    #include<iostream.h> int main() { int a, b; cout<<"Please enter two numbers: "; cin>>a>>b; cout<<a<<"+"<<b<<"="<<a+b<<endl; return 0; }
    // 1.2 直角三角形已知两直角边,求斜边长
    
    #include<iostream.h>
    #include<math.h>
    int main()
    {
        double a,b;
        cout<<"请输入两直角边的长: 
    ";
        cin >>a>>b;
        cout.precision(4);
        cout<<"斜边长为:"<<sqrt(a*a+b*b) <<endl;
        return 0;
    }
    // 1.3时间转换(把秒转换为标准时间)
    // 例:3600 s 转化为 01:00:00
    

    //(1)李韡
    #include<iostream.h> int main() { int sum, hour, min, second; cout<<"Please enter a data:"<<endl; cin>>sum; if(sum<0) cout<<"error."; else { hour = sum/3600; min = (sum-hour*3600)/60; second = (sum-hour*3600-min*60); } output(hour); output(min); output(second); cout<<" "; return 0; } void output(int a) { if(a<10) cout<<"0"<<a<<":"; else cout<<a<<":"; } //(2)査俊毅 #include<iostream.h> int main() { int time; int H,M,S; there: cout<<"请输入秒数: "; cin>>time; if(time<0||time>86400) { cout<<"输入有误!a "; goto there; } else { H=time/3600; M=(time%3600)/60; S=(time%3600)%60; cout<<"现在的时间是: "<<H<<""<<M<<""<<S<<""<<endl; } return 0; }
    //  1.4求三个实数的和并将其输出,再输出将其四舍五入之后的值
    //  翁世杰  
    
    #include<iostream.h>
    int main(void)
    {    
        float x=0, y=0, z=0;
        float sum=0, temp;
        cout<<"Please enter three real number:"<<endl;
        cin>>x>>y>>z;
        sum = x + y + z;
        cout<<"sum="<<sum;
        cout<<"
    After rounding the result:
    ";    
        temp=sum-int(sum);                            
        if(temp>=0.5)    cout<<"sum="<<int(sum)+1<<endl;        
        else            cout<<"sum="<<int(sum)<<endl;
        return 0;
    }
    // 1.5.1 成绩转换成等级(if else)
    //(输入一个100分制的成绩,将其转换为相对应的等级,90以上A,80以上B,70以上C,60以上D,60一下E)
    // 李韡 #include<iostream.h> int main() { int score; cout<<"Please enter your score: "; while(cin>>score) { if(score > 100 || scor e< 0) cout<<"Wrong Data. "; else { if(score>=90) cout<<"Grade: A"; else if(score>=80) cout<<"Grade: B"; else if(score>=70) cout<<"Grade: C"; else if(score>=60) cout<<"Grade: D"; else cout<<"Grade: E"; cout<<endl; } } return 0; }
    // 1.5.2 成绩转换成等级(switch)
    //(输入一个100分制的成绩,将其转换为相对应的等级,90以上A,80以上B,70以上C,60以上D,60一下E)
    // 李韡
    #include<iostream.h>
    int main()
    {
        int score;
        cout<<"Please enter your score:  ";
        while( cin>>score )
        {    
            if( score>100 || score<0 ) 
                cout<<"Wrong Data.
    ";
            else
            {
                score/=10;
                switch(score)
                {
                    case 10: 
                    case  9:  cout<<"Grade: A
    "; break; 
                    case  8:  cout<<"Grade: B
    "; break; 
                    case  7:  cout<<"Grade: C
    "; break; 
                    case  6:  cout<<"Grade: D
    "; break; 
                    default:  cout<<"Grade: E
    "; break;
                }
            }
        }
        return 0;
    }
    //1.6 求1000以内的所有完数
    //翁世杰
    #include<iostream.h> #define N 1000 int main() { int i, j, sum; for(i=1; i<=N; i++) { for(j=1,sum=0;j<=i/2;j++)//一个数的因子至少是其二分之一 { if(i%j==0) sum+=j; } if(sum==i) cout<<i<<endl; } return 0; }
    //1.7 解二元一次方程组
    //张世昊
    #include<iostream.h>
    #include<math.h>
    void main()
    {
        int A,B,C;
        int a,b;
        cout<<"请输入一元二次方程的三个系数"<<endl;
        double X,Y_1,Y_2;
        cin>>A>>B>>C;
        X=B*B-4*A*C;
        if(A==0){
            cout<<"输入值不满足一元二次方程"<<endl;}
        else
        {
            if(X<0)    
                cout<<"此方程无解"<<endl;
            else
            {
                a=2*A;
                b=-B;
                Y_1=(b-sqrt(X))/a;
                Y_2=(b+sqrt(X))/a;
                if(Y_1==Y_2)
                     cout<<"此方程的两个解相等为:"<<Y_1<<endl;
                else
                    cout<<"次方程有两个不相等的解分别为:"<<Y_1<<"   "<<Y_2<<endl;
            }
        }
    }
     
    // 1.8 打印九九乘法表
    // 李韡
    #include<iostream.h>
    #include <iomanip.h>
    int main()
    {
        int i,j;
        cout<<"*";
        for(i=1;i<10;++i)
            cout<<setw(4)<<i;
        cout<<"
    -------------------------------------
    ";
        for(i=1;i<10;++i)
        {
            cout<<"
    "<<i;
            for(j=1;j<10;++j )    
                cout<<setw(4)<<j*i;
        }
        cout<<endl;    
        return 0;
    }
    // 1.9 求两个数的最大公约数
    // 李韡
    #include<iostream.h>
    int main()
    {
        int a, b, c, t;
        cout<<"Please enter two number:"<<endl;
        cin>>a>>b;
        if(a<b)
        {
            t=a; 
            a=b; 
            b=t;
        }
        c=a%b;
        while(c!=0)                            
        {
            a=b;b=c;c=a%b;
        }
            cout<<"The gcd of the two numbers is: "<<b<<endl;
        return 0;
    }
    // 1.10 用*输出三角形
    // 李韡
    
    #include<iostream.h>
    int main()
    {
        int i, j, line;
        cout<<"请输入要打印出的三角形的行数:"<<endl;
        cin>>line;
        for(i=1; i<=line; i++)        //控制总的行数
        {
            for(j=i; j<line; j++)    //控制空格数
            {
                cout<<" ";
            }
            for(j=1; j<=i; j++)        //控制*的个数
                cout<<"* ";
            cout<<endl;
        }
        return 0;
    }
    //1.11 输出平行四边形
    // 张世昊
    #include<iostream.h>
    int main()
    {
        int i,j,k,t;
        int a,b;
        cout<<"输入平行四边形的高度a=";
        cin>>a;    
        cout<<"输入平行四边形的长度b=";
        cin>>b;
        t=a+1;
        for(i=1;i<=a;i++)
        {
            for(k=t;k>=1;k--)
                cout<<" ";
            t--;
            for(j=1;j<=b;j++)
                cout<<"+ ";
            cout<<"
    ";
        }
    }
    //1.12 输出ABA菱形
    // 白磊   
    #include<iostream.h>
    void main()
    {
        int i,j,k;
        cout<<"请输入2~20的一个数:
    ";
        cin>>k;    
    
        for(j=1;j<=k+1;j++)//输出第一个A
        {
            if(j==k+1)
            break;
            cout<<" ";
        }
        cout<<"A 
    ";
        for(i=1;i<=k;i++)
        {    
            for(j=1;j<=k-i;j++)//控制输出空格
                cout<<" ";
            cout<<"A ";
            for(j=1;j<=i-1;j++)
                cout<<"B ";
            cout<<"A"<<"
    ";
        }
        for(i=1;i<=k-1;i++)
        {
            for(j=1;j<=i;j++)
                cout<<" ";
            cout<<"A ";
            for(j=1;j<=k-i-1;j++)
                cout<<"B ";
        cout<<"A";
        cout<<"
    ";
        }
        for(i=1;i<=k;i++)//输出最后一个A
            cout<<" ";
        cout<<"A"<<endl;
    }
    //1.13 杨辉三角
    // 李韡
    /*
    用二维数组来做,
    规律是每一列的第一项和最后一项都是1,
    也就对应我们的二维数组(a[i][j])中的i=j或者j=0的时候的情况。
    中间的数字,
    a[i][j] = a[i-1][j-1] +a[i-1][j]
    */
    /*
                         1
                       1   1
                     1   2   1
                   1   3   3   1
                 1   4   6   4   1
               1   5  10  10   5   1
             1   6  15  20  15   6   1
           1   7  21  35  35  21   7   1
         1   8  28  56  70  56  28   8   1 
       1   9  36  84 126 126  84  36   9   1
    Press any key to continue
    */
    /*
    改进,改为等腰三角形的杨辉三角,,
    注意了空格的输出
    */
    #include<iostream.h>
    #include<iomanip.h>
    int main()
    {    
        int a[10][10]={0,0};
        int b[10][10]={0,0};
        int i, j, k, t;
        for (i=0;i<10;i++)  
        {   
            for(k=i;k<18-i;k++)                
                cout<<" ";
            for (j=i;j>=0;j--)  
            {  
                if (i==j || j==0) 
                    a[i][j]=1;
                else
                    a[i][j] = a[i-1][j-1]+a[i-1][j];
            }  
            for (j=i;j>=0;j--)  
                cout<<setw(4)<<a[i][j];                //域宽
            cout<<endl;
        }  
        return 0;
    }
    // 1.13 杨辉三角
    // 查俊毅
    #include<iostream.h>
    #include<iomanip.h>
    int zjy(int m)
    {    
        int i,k;
        k=1;
        for(i=1;i<=m;i++)
            k=i*k;
        return k;
    }
    int main()
    
    {    int a,b,n;
        cout<<"请输入杨辉三角的行数:";
        cin>>n;
        for(a=0;a<=n-1;a++)
        {
            for(b=1;b<=n-a-1;b++)
                cout<<"    ";
            for(b=0;b<=a;b++)
                cout<<setw(8)<<zjy(a)/zjy(b)/zjy(a-b);
            cout<<endl;
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Lee-geeker/p/3150099.html
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