hdu_1170_Balloon Comes!_水之

水..

http://acm.hdu.edu.cn/showproblem.php?pid=1170

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11465    Accepted Submission(s): 3998

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

3-120.50

 

#include<iostream>
#include<stdio.h>
using namespace std;
int main()
{
    double wa;
    int t,a,b;
    char c;
    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=='/')
        {
            wa=(double)a/b;
            if((wa-(int)wa)==0)
                cout<<a/b<<endl;
            else
                printf("%.2f\n",wa);
        }
    }
    return 0;
}