1410242114-hd-Balloon Comes!

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20169    Accepted Submission(s): 7597

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
   
   
   
   

题目大意

       给定一个符号和两个整数，根据符号对这两个数进行运算，如果存在小数的话就保留两位小数

代码

#include<stdio.h>
int main()
{
	int t;
	int a,b;
	char c;
	int i,j;
	scanf("%d",&t);
	getchar();
	while(t--)
	{
		scanf("%c%d%d",&c,&a,&b);
		getchar();
		switch(c)
		{
			case'+':printf("%d\n",a+b);break;
		    case'-':printf("%d\n",a-b);break;
		    case'*':printf("%d\n",a*b);break;
		    case'/':{//如果得到的不是整数的话就需要保留2位小数。 
				         if(a%b==0)
			                 printf("%d\n",a/b);
						 else
						     printf("%.2lf\n",a/(b*1.0));
					 }break;
		}
	}
	return 0;
}