HDU 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

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

题目大意：

先输入一个数，表示接下来有几组测试数据。测试数据会给你一个操作符（+，，，/ -表示加法，减法，乘法，除法）和两个正整数，你的任务是输出结果。

c++

#include <iostream>#include<cstdio>#include<cstring>using namespace std;int main(){    char a;    int b,c,d;    double e,f;    cin>>b;    while(b--)    {        cin>>a>>c>>d;        if(a=='+')      //判断操作符是什么        {            e=c+d;            printf("%.0lf\n",e);        }        else if(a=='-')        {            e=c-d;            printf("%.0lf\n",e);        }        else if(a=='*')        {            e=c*d;            printf("%.0lf\n",e);        }        else        {            e=c*1.0/d;            if(c%d==0)            {                printf("%.0lf\n",e);            }            else            {                printf("%.2lf\n",e);            }        }    }    return 0;}