Balloon Comes!

浅浅的花香味﹌ 2022-08-10 05:50 152阅读 0赞

## Balloon Comes! ##

##### Time Limit: 1000ms Memory limit: 65535K 有疑问？点这里^\_^ #####

## 题目描述 ##

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!

## 输入 ##

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.

## 输出 ##

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.

## 示例输入 ##

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

## 示例输出 ##

3
    -1
    2
    0.50

## 提示 ##

[hdoj1170][] 有链接提示的题目请先去链接处提交程序，AC后提交到SDUTOJ中，以便查询存档。

## 来源 ##

HDOJ lcy

#include <stdio.h>   
    int main()   
    {   
    int i,j,n,m,b;  
    char a;  
    scanf("%d\n",&n);  
    for(i=0;i<n;i++)  
    {  
        scanf(" %c %d %d",&a,&m,&b);  
        if(a=='+')  
            printf("%d\n",m+b);  
        else if(a=='-')  
            printf("%d\n",m-b);  
        else if(a=='*')  
            printf("%d\n",b*m);  
        else if(a=='/')  
            printf("%.2lf\n",1.0*m/b);  
      
    }  
    }

[hdoj1170]: http://acm.hdu.edu.cn/showproblem.php?pid=1170