增加功能的四则运算

标签：oid   case   想法   循环   switch   use   ase   include   sig   

一：想法：1.支持真分数，用到数学函数show（）

                   2.30道运算所以用到循环函数 i+1

                   3.保证没有重复的算式，所以用到for语句。

二：程序代码：#include<stdio.h>
#include<Windows.h>
#include<time.h>
void main()
 {
      int a, b, c, d,i,m,n;
      float p, q;
     srand(unsigned( time(NULL)));
      for (i = 0; i < 30; i++)
    {
        a = rand() % 100;
        b = rand() % 100;
         c = rand() % 5;
         switch (c)
         {
        case 0: printf("%d + %d =\n", a, b);
             break;
        case 1:if (a < b)
         {
                    d = a;
                   a = b;
           }
                printf("%d - %d =\n", a, b);
               break;
         case 2:printf("%d * %d =\n", a, b);
             break;
         case 3:
            while (b == 0)
            {
                b = rand() % 100;
             }
            printf("%d / %d =\n", a, b);
            break;
        case 4:
             a = rand() % 100;
             b = rand() % 100;
             m = rand() % 100;
            n = rand() % 100;
             c = rand() %4 ;
             while (b == 0)
            {
                b = rand() % 100;
            }
           while (n == 0)
          {
                n = rand() % 100;
             }
             if (a > b)
            {
                d = a;
                a = b;
                b = d;
            }
            if (m > n)
             {
                 d = m;
                m = n;
                 n = d;
             }
            switch (c)
             {
           case 0:printf("%d/%d + %d/%d =\n", a, b, m, n);
                 break;
            case 1:
                 p = (float)a / (float)b;
               q = (float)m / (float)n;
                 if (p < q)
             {
                   d = a;
                     a = m;
                     m = d;
                     d = b;
                     b = n;
                     n = d;
                 }
                printf("%d/%d - %d/%d =\n", a,b,m,n);
                 break;
            case 2:printf("%d/%d * %d/%d =\n", a, b, m, n);
                 break;
             case 3:
                a = rand() % 100;
                b = rand() % 100;
                 m = rand() % 100;
                n = rand() % 100;
                 c = rand() % 4;
                 while (m == 0)
                 {
                     m = rand() % 100;
                }
                 printf("%d/%d / %d/%d =\n", a, b, m, n);
                break;
             }
         }
     }
     system("pause");
 }

三：运行结果截图：

增加功能的四则运算

标签：oid   case   想法   循环   switch   use   ase   include   sig   

原文地址：https://www.cnblogs.com/sunyangkai/p/9751852.html