Balloon Comes!

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

 

#include <stdio.h>

int main(){
    int T;
    char c;
    int a;
    int b;
    
    scanf("%d",&T);
    
    while(T--){
        getchar();
        
        scanf("%c%d%d",&c,&a,&b);
        
        if(c=='+')
            printf("%d\n",a+b);
            
        else if(c=='-')
            printf("%d\n",a-b);
            
        else if(c=='*')
            printf("%d\n",a*b);
            
        else if(c=='/'){
            if(a%b!=0)
                printf("%.2lf\n",(double)a/b);
                
            else
                printf("%d\n",a/b);
        }
    }
    return 0;
}