C++自主测试题目

下面是题目

后面有代码

1.键盘输入3个整数a,b,c值，求一元二次方程a*X∧2+b*X+c=0(a≠0)的根，结果保留两位小树。

2.编写一个口令输入程序，让用户不停输入口令，直到输对为止，假设口令为100。

3.若一个数恰好等于除它本身外的所有因子之和，则这个数成为完数，例：6点因数是1，2，3。且6=1+2+3所以6是完数，求100以内的所有完数，并输出。

4.编程从键盘输入10个数，求它们的方差。

5.编程将递增数列10，20，30，40，50，60，70，80，90，100保存到数组中，再从键盘输入一个整数，将它插入该数列中，使之扔为一个递增数列。

6.键盘输入一个M*N的二位数组，求该数组各行的平均值，将结果放到一个数组中，并输出。

7.编写函数判断一个整数是否为回文数。主函数中调用该函数，输出1000到9000之间的所有回文数，，每行输出十个数字。回文是指正读和倒读都一样的数字。如98789。

8.编写函数power()用来求n∧k的值。在主函数中输入两个正整数n和k，调用函数power，求1∧k+2∧k+…+n∧k的值并输出结果。

9.编程求C(n,m)=m！/n！*(m-n)！,要求在主函数中输入自然数m,n的值并输入结果。若输入错误(即输入的不是自然数)，提示错误信息。

10.编写函数输入月份，输出该月的英文名，要求用指针数组实现。

(附加题)11.     加密程序:由键盘输入明文，通过加密程序转换成密文并输出到屏幕上。 算法：明文中的字母转换成其后的第4个字母,例如，A变成E(a变成e)，Z变成D,非字母字符不变；同时将密文每两个字符之间插入一个空格。例如，China转换成密文为G l m r e。要求：在函数change中完成字母转换，在函数insert中完成增加空格，用指针传递参数。

附加题别问了 我也忘了怎么写的了

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#include <iostream>
#include<stdlib.h>
#include<stdio.h>
#include<math.h>
using namespace std;

/******************************C语言测试*******************************************************/
void test001()
{
    int a, b, c;
    double decide, x1, x2, m, n;
    cout << "Please key in a，b，c";
    cin >> a >> b >> c;

    decide = (double)b * (double)b - 4 * (double)a * (double) c;
    if (decide < 0)
    {
        printf("The number is roung \n");
    }
    else
    {
        m = -b / (2 * a);
        n = sqrt(decide) / (2 *(double) a);
        x1 = m + n;
        x2 = m - n;
        system("color 02");
        printf("The answer is:\nx1=%.2f\nx2=%.2f\n", x1, x2);
    }
}

void test002()
{
    int b;
    cout << "Please key in the answer\n";
    while (1)
    {
        cin >> b;
        if (b == 100)
        {
            break;
        }
        else
        {
            system("color 02");
            cout << " 输入错误，再来一次\n";
        }
    }
    system("color 02");
    cout << "我的妈呀 你终于输入正确了";
}

void test003()
{
    int i, j, sum = 0;

    for (i = 2; i <= 100; i++)
    {
        for (j = 1; j < i; j++)
        {
            if (i % j == 0)
            {
                sum = sum + j;
            }
        }
        if (sum == i)
        {
            printf("%d \n", i);
        }

        sum = 0;
    }
}

void test004()
{
    int a[11], i, m,sum,average,ans,answer;
    sum = 0;
    ans = 0;
    for (i = 0; i < 10; i++)
    {
        cin >> a[i];
    }
    for (i = 0; i < 10; i++)
    {
        sum= sum+ a[i];
    }
    average = sum / 10;
    cout << sum <<endl<< average<<endl;
    for (i = 0; i < 10; i++)
    {
        ans = ans + pow((a[i] - average), 2);
    }
      answer = ans / 10;
    cout << answer;
}

void test005()
{
    int m, n, v,i,c;
    int a[15] = { 10,20,30,40,50,60,70,80,90,100 };
    int j[15] = { 10,20,30,40,50,60,70,80,90,100 };
    cin >> v;
    for (i = 0; i < 10; i++)
    {
        if (v > a[i] && v < a[i+1])
        {
           m = i + 1;
           break;
        }
     }

    n = m;
    for (n; n<14; n++)
    {
        a[n+1] = j[n];
    }
    a[m] = v;
    for (c = 0; c <=12; c++)
    {
        cout << a[c]<<endl;
    }
}

void test006()
{

}

void test007()
{
    int a, i = 0;
        for(a=1000;a<9001;a++)
        {
            if(a/1000%10==a/1%10 && a/100%10==a/10%10)
            {
                printf("%5d", a);
                i++;
                if(i==9)
                {
                    printf("\n");
                    i = 0;
                }
            }
        }
}

void test008(int n,int k)
{
    int m,sum=0;
    for (m = 1; m <= n; m++)
    {
        sum = sum + pow(m, k);
    }
    cout << sum;
}

 int jiecheng( int m)
{
    int Plus=1;
    for (int i = 1; i <= m;i++)
    {
        Plus= Plus*i;
    }
    return Plus;
}

void test009()
{
    double answer; int n, m;
    cout << "Please key in two number" << endl;
    cin >> n >> m;
    if (n == int(n) && m == int(m) && m > 0 && n > 0)
    {
        answer =(double) jiecheng(m) / (double)(jiecheng(n) * (double)jiecheng(m - n));
        cout << answer;
    }
    else
    {
        cout << "error";
    }
}

void test010()
{
    int i;
    const char* p[12] = { "January","February","March","April","May","June","July","August","September","October","November","December" };
    printf("please input your number of month:");
    cin >> i;
    printf("\n");
    //    i = i - 1;
//    printf("The month is:%s\n", *p+i);
    printf("The month is:%s\n", p[i-1]);
}

void test010_SECOND()
{
    int i;
     char p[12][30] = { "January","February","March","April","May","June","July","August","September","October","November","December" };
    printf("please input your number of month:");
    cin >> i;
    char* a;
    a = p[12];
    printf("\n");
    printf("The month is:%s\n", *(p+i - 1));
}

void test006()

{
    int a, b, c = 0, d;
    int column, row;
    int sum[100] = { 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 };
    cout << "输入二维动态数组的行数和列数" << endl;
    cin >> row >> column;
    int n1, i;
    int** array;
    array = (int**)malloc(sizeof(int*) * row);
    for (int i = 0; i != row; i++)
    array[i] = (int*)malloc(sizeof(int) * column);
    int* average;
    n1 = row;
    average = (int*)malloc(n1 * sizeof(int));//动态一维数组
    cout << "输入二维数组"<<endl;
    for (int j = 0; j != row; j++)
    {
        for (int k = 0; k != column; k++)
        {
            cin >> array[j][k];
        }
    }

    for (int j = 0; j != row; j++)
    {
        for (int k = 0; k != column; k++)
        {
            cout << array[j][k];
        }
    }

    /*
    for (a = 0; a <= row; a++)
    {
        for (b = 0; b < column; b++)
        {
            sum[a] = sum[a] + array[a][b];
        }
        average[a] = sum[a] / row;
    }

    for (c = 0; c <= row; c++)
    {
        cout << average[c];
    }

    */

    //释放空间
    for (int i = 0; i != row; i++)
    {
        free(array[i]);
    }
    free(array);
    free(average);
}

int main()
{
    test001();
    test002();
    test003();
    test004();
    test005();
    test006();
    test007();
    int a, b;
    cin >> a >> b;
    test008(a,b);
    test009();
    test010();
    return 0;

}

原文地址：https://www.cnblogs.com/Loving-Q/p/11963770.html