上题:

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1611 Accepted Submission(s): 533

Problem Description
The contest starts now! How excited it is to see balloons floating around. You, one of the best programmers in HDU, can get a very beautiful balloon if only you have solved the very very very... easy problem.
Give you an operator (+,-,*, / --denoting addition, subtraction, multiplication, division respectively) and two positive integers, your task is to output the result.
Is it very easy?
Come on, guy! PLMM will send you a beautiful Balloon right now!
Good Luck!
 

Input
Input contains multiple test cases. The first line of the input is a single integer T (0<T<1000) which is the number of test cases. T test cases follow. Each test case contains a char C (+,-,*, /) and two integers A and B(0<A,B<10000).Of course, we all know that A and B are operands and C is an operator.
 

Output
For each case, print the operation result. The result should be rounded to 2 decimal places If and only if it is not an integer.
 

Sample Input
4
+ 1 2
- 1 2
* 1 2
/ 1 2
 

Sample Output
3
-1
2
0.50

贴代码:

#include<stdio.h>
int main()
{
    char ch;
    int a,b;
    int i,n;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%c%d%d",&ch,&a,&b);
        switch(ch)
        {
            case '+': printf("%d\n",a+b);break;
            case '-': printf("%d\n",a-b);break;
            case '*': printf("%d\n",a*b);break;
            case '/':
                if(a%b==0)
            {printf("%d\n",a/b);break;}
                else
            {
                    printf("%.2f\n",(float)a/b);
                    break;
            }
        }
    }

    return 0;
}

问题来了:我输入
4
+ 1 2
- 2 1
然后程序就结束了,为什么????