hdu 1170 Balloon Comes!

小咪咪 2022-05-31 09:24 167阅读 0赞

# Balloon Comes! #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 32966 Accepted Submission(s): 12430**

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

题目解析：

非常简单的题目，不多说，看代码

代码：

#include <bits/stdc++.h>
    using namespace std;
    int main()
    {
        //freopen("in.txt","r",stdin);
        int T;
        cin>>T;
        char c;
        int a,b;
        while(T--)
        {
            double ans=0;
            cin>>c>>a>>b;
            if(c=='+')
                ans=a+b;
            if(c=='-')
                ans=a-b;
            if(c=='*')
                ans=a*b;
            if(c=='/')
                ans=a*1.0/b;
            if(ans==(int)ans)
                cout<<(int)ans<<endl;
            else
                printf("%.2f\n",ans);
        }
        return 0;
    }