ACM HDU 1170 Balloon Comes! （完全的水题，加减乘除）

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10546    Accepted Submission(s): 3623

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

 

Author

lcy

 

水题，注意

The result should be rounded to 2 decimal places If and only if it is not an integer.就是做除法如果整除要输出整数，因为这个WR了

#include<stdio.h>
#include<iostream>
using namespace std;
int main()
{
    int T,a,b;
    char ch;
    scanf("%d",&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);//这里注意，看清题目意思 
            else 
               printf("%.2f\n",(float)a/b);
        }    
    }    
    return 0;
}