hdoj 1170 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.

 

SampleInput

4
+ 1 2
- 1 2
* 1 2
/ 1 2

 

SampleOutput

3
-1
2
0.50

如果不是整数
结果应该四舍五入为小数点后2位

#include <iostream>
#include <stdio.h>
using namespace std;
int main() {
    int t,a,b;
    char ch;
    cin>>t;
    while(t--){
        cin>>ch>>a>>b;
        if(ch=='+') printf("%d\n",a+b);
        else if(ch=='-') printf("%d\n",a-b);
        else if(ch=='*') printf("%d\n",a*b);
        else{
            if(a%b==0) printf("%d\n",a/b);
            else printf("%.2f\n",(float)a/b);
        }
    }
    return 0;
}