HDU 1170 Balloon Comes!（计算表达式）

                                     Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25560    Accepted Submission(s): 9664

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

题意：没什么好说的。。。。

AC代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
typedef long long LL;
using namespace std;
int main()
{
    int n;
   scanf("%d",&n);
   getchar();
   while (n--)
   {
      char c;
      int a,b;
      scanf("%c%d%d",&c,&a,&b);
      getchar();  //记得把回车消除，不然会WA 
      switch (c)
      {
      case '+':
      printf("%d\n",a+b);break;
      case '-':
      printf("%d\n",a-b);break;
      case '*':
      printf("%d\n",a*b);break;
      case '/':
      a%b==0?printf("%d\n",a/b):printf("%.2lf\n",(double)a/(double)b);break;
      
      }
   }
         
   return 0;
}