1．程序随机产生两位数以内的n（如10）道算术题，要学生回答。
  2．n道题做完后，程序给出评语：
（1）做对9题以上：优秀；
（2）做对7或8道题：良好；
（3）做对5或6道题：不错；
（4）其他：还要努力。
  3．学生做完一轮后，不用退出，可以进入下一轮。

运算符只有+-*/？

#include<stdio.h>
#include<stdlib.h>
#include<time.h>
int  main()
{
    srand(time(0));
    printf("输入出题个数,以0结束。\n");
    int n, cj = 0, a, b, c, jg;
    scanf("%d", &n);
    while (n)
    {
        for (int i = 0; i < n; i++)
        {
            a = rand() % 90 + 9;
            b = rand() % 90 + 9;
            c = rand() % 4 + 1;
            switch (c)
            {
            case 1:
                printf("%2d + %2d=", a, b);
                scanf("%d", &jg);
                if (jg == a + b)cj++;
                break;
            case 2:
                printf("%2d - %2d=", a, b);
                scanf("%d", &jg);
                if (jg == a - b)cj++;
                break;
            case 3:
                printf("%2d * %2d=", a, b);
                scanf("%d", &jg);
                if (jg == a * b)cj++;
                break;
            case 4:
                printf("%2d / %2d=", a, b);
                scanf("%d", &jg);
                if (jg == a / b)cj++;
                break;
            }
            printf("\n");
        }
        switch (cj * 100 / n)
        {
        case 90:
        case 100:
            printf("正确率%6.2f%%优秀\n", cj * 100*1.0 / n);
            break;
        case 70:
        case 80:
            printf("正确率%6.2f%%良好\n", cj * 100 *1.0/ n);
            break;
        case 50:
        case 60:
            printf("正确率%6.2f%%不错\n", cj * 100 *1.0/ n);
            break;
        default :
            printf("正确率%6.2f%%还须努力\n", cj * 100*1.0 / n);
            break;
        }
        cj=0;
        printf("输入出题个数,以0结束.\n");
        scanf("%d", &n);
    }
    return 0;
}只能进行加减乘除运算

[此贴子已经被作者于2018-5-21 17:17编辑过]

多谢多谢  我一会运行一下

应该是加减乘除吧

[此贴子已经被作者于2018-5-21 20:33编辑过]

#include<stdio.h>
#include<stdlib.h>
#include<time.h>
double suiji()
{
    int a,b,k;
    double sum;
    a=rand()%100;
    b=rand()%100;
    k=rand()%4;
    printf("%2d",a);
    switch(k)
    {
        case 0:
            printf("+");
            sum=a+b;
            break;
        case 1:
            printf("-");
            sum=a-b;
            break;
        case 2:
            printf("*");
            sum=a*b;
            break;
        case 3:
            while(1)
            {
                if(b==0)
                {
                b=rand()%100;
                }
                else
                break;
            }
            printf("/");
            sum=a*1.0/b;
            a=sum*100+5;
            sum=a/100.0;
            break;
        default:
            break;
    }
    printf("%2d\n",b);
    return sum;
}
void main()
{
    printf("请输入题目数量\n") ;
    srand((unsigned) time(NULL));
    while(1)
    {
        int n;
        scanf("%d",&n);
        double a[n],k=0;
        char m;
        double c;
        for(int i=0;i<n;i++)
        {
            printf("第%2d题",i+1);
            a[i]=suiji();
        }
        printf("请按题号输入每个题目的答案用空格隔开\n");
        for(int i=0;i<n;i++)
        {
            scanf("%lf",&c);
            if(c==a[i])
            k++;
        }
        if(k>(n*8+5)/10)
            printf("优秀");
        else if(k>(n*6+5)/10)
            printf("良好");
        else if(k>(n*4+5)/10)
            printf("不错");
        else
            printf("继续努力");
        k=0;
        for(int i=0;i<n;i++)
            printf("第%d题答案为%.2lf\n",i+1,a[i]);
        printf("输入'N'退出答题；继续答题请输入'Y' ");
        scanf("%c",&m);
        scanf("%c",&m);
        if(m=='N')
        {
            break;
        }
    }   
}
大体和楼上差不多
只是避免被除数为0 的情况

重复了，忽略这楼

运行出来了 可以给个联系方式吗 说好的有偿  不能让你白编写

十分感谢 你可以加一下我的微信  1205558907   标题都说了 有偿的  不会让你白写程序的

[此贴子已经被作者于2018-5-21 23:55编辑过]

都是学习不用有偿，健壮性还不行，你自己改一下。