用C语言做一个简单的计算器程序要求如图

Question

如题所述

Answer 1

#include <stdio.h>
int main()
{
     char judge,oper;
     double a, b,result;
     printf("是
否需要计算，输入Y或y表示需要，输入N或n表示不需要\n");
     judge = getchar();
     while (judge != 'N' && judge != 'n')
     {
          if (judge == 'Y' || judge == 'y')
          {
               printf("请输入需要计算的两个数，空格分开\n");
               scanf("%lf %lf", &a, &b);
               while (getchar() != '\n');
               printf("请输入操作符\n");
               oper = getchar();
               switch (oper)
               {
               case '+':
                    result = a + b;
                    break;
               case '-':
                    result = a - b;
                    break;
               case '*':
                    result = a*b;
                    break;
               case '/':
                    if (b == 0)
                         break;
                    else
                    {
                         result = a / b;
                         break;
                    }
               default:
                    printf("错误的操作符\n");
                    break;
               }
               if (b == 0 && oper == '/')
               {
                    printf("除法中除数不可为0\n");
               }
               else if (oper == '+' || oper == '-' || oper == '*' || oper == '/')
               {
    
                    printf("结果约为%.2lf\n", result);
               }
               printf("是否需要计算，输入Y或y表示需要，输入N或n表示不需要\n");
               while (getchar() != '\n');
               judge = getchar();
          }
          else
          {
               printf("您输入的不是Y,y,N,n中的字符，请重新输入，Y或y表示需要计算，N或n表示不需要\n");
               while (getchar() != '\n');
               judge = getchar();
          }
     }
     while (getchar() != '\n');
     printf("程序结束，ENTER关闭程序\n");
     getchar();
     return 0;
}

格式可能有点乱，粘贴过来后格式丢失了