hdu 1170 Balloon Comes!

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26455    Accepted Submission(s): 10055

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

 

分析：简单题，只要注意除法的时候，如果能整除则直接输出就行，否则保留两位小数

#include <iostream>
#include <cstdio>
using namespace std;

int main(){
    char c;
    int t, a, b;
    cin >> t;
    while(t--){
        cin >> c >> a >> b;
        if(c == '+')
            cout << (a + b) << endl;
        else if(c == '-')
            cout << (a - b) << endl;
        else if(c == '*')
            cout << (a * b) << endl;
        else if(c == '/'){
            if(a % b == 0)
                cout << (a / b) << endl;
            else
                printf("%.2f
", (float)a / b);
        }        
    }
    return 0;
}