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!=)
                 printf("%.2lf\n",(double)a/b);

             else
                 printf("%d\n",a/b);
         }
     }
     return ;
 }