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Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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Total Submission(s): 26449    Accepted Submission(s): 10051


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


还是一道水题，但是有几点注意的

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#define LL long long
using namespace std;
char str;
int n,m;
int main()
{
  int t;
  scanf("%d",&t);
  while(t--)
  {
    //scanf("%s %lf %lf",str,&n,&m);
    cin>>str>>n>>m;
    if(str=='+')  printf("%d\n",n+m);
    if(str=='-')  printf("%d\n",n-m);
    if(str=='*')  printf("%d\n",n*m);
    if(str=='/')
    {   // double 型不能用 % 取余 
      if(n%m==0)  printf("%d\n",n/m); // 能整除就不输出浮点型，这里一直 WA  
      else printf("%.2lf\n",(double)(n+0.0)/m);
    }
  }
  return 0;
}