#include"stdio.h"
main()
{
    int n,a,i,b;
    char c;
    scanf("%d",&n);
    for(i=n;i>0;i--)
    {
        scanf("%c",&c);
        switch(c)
        {
            case'+':
            {
                scanf("%d %d",&a,&b);
                printf("%d\n",a+b);
                break;
            }
            case'-':
            {
                scanf("%d %d",&a,&b);
                printf("%d\n",a-b);
                break;
            }
            case'*':
            {
                scanf("%d %d",&a,&b);
                printf("%d\n",a*b);
                break;
            }
            case'/':
            {
                scanf("%d %d",&a,&b);
                printf("%.2f\n",(a*1.0)/b);
                break;
            }
        }   
    }
}

输入数据如下 为什么到第二次就结束了?
4
+ 1 2
- 1 2
* 1 2
/ 1 2

回车也被读入了。
在scanf("%c",&c);前面加fflush(stdin);

好像没什么效果哦。不过还是挺感谢的。。。

#include "stdio.h"
main()
{
    int n,a,i,b;
    char c;
    scanf("%d",&n);
    for(i=n;i>0;i--)
    {
        printf("i=%d\n",i);
        fflush(stdin);
        scanf("%c",&c);
        switch(c)
        {
            case'+':
            {
                scanf("%d %d",&a,&b);
                printf("%d\n",a+b);
                break;
            }
            case'-':
            {
                scanf("%d %d",&a,&b);
                printf("%d\n",a-b);
                break;
            }
            case'*':
            {
                scanf("%d %d",&a,&b);
                printf("%d\n",a*b);
                break;
            }
            case'/':
            {
                scanf("%d %d",&a,&b);
                printf("%.2f\n",(a*1.0)/b);
                break;
            }
            default:
                printf("wrong c = %c", c);
                break;
        }   
    }
}

n给一个确定的数字吧

http://acm.hdu.
这样设计？为什么错？我实在搞不懂？求教求教
#include "stdio.h"
main()
{
    int n,a,i,b;
    char c;
    scanf("%d",&n);
    for(i=n;i>0;i--)
    {
        fflush(stdin);
        scanf("%c",&c);
        switch(c)
        {
            case'+':
            {
                scanf("%d %d",&a,&b);
                printf("%d\n",a+b);
                break;
            }
            case'-':
            {
                scanf("%d %d",&a,&b);
                printf("%d\n",a-b);
                break;
            }
            case'*':
            {
                scanf("%d %d",&a,&b);
                printf("%d\n",a*b);
                break;
            }
            case'/':
            {
                scanf("%d %d",&a,&b);
                printf("%.2f\n",(a*1.0)/b);
                break;
            }
            default:
                break;
        }   
    }
}

你原来的代码
输入：4  回车 + 空格1空格 2 回车 -空格 1空格 2 回车 *空格 1空格。。。
      n=4
 1       c=回车switch不通过第一次循环结束
2           c=+，switch要求输入两个数所以空格跳过a=1，b=2，循环结束
3                           c=回车，switch不通过此次循环结束
4                                c=-，switch结束要求输入两个数a=1，b=2
 4次循环结束程序结束。

题目完全不同嘛

#include<conio.h>
 for(i=n;i>0;i--)
    {
      
        getchar();//添加用来吃掉回车符
        scanf("%c",&c);
        switch(c)
       ...

求这道题的解题思路。。。