Balloon Comes!

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20582 Accepted Submission(s): 7781

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
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    解题思路：注意考虑测试数据输入，回车键作为数据键入数据处理
   
   
   
   
   
   
   
   

    
    
    
    源代码： #include <stdio.h> #include <string.h> #include <stdlib.h> int main() {   int a,b,t;   char c;   scanf("%d",&t);   while(t--)   {       getchar();//每次输入测试数据后，注意换行字符      scanf("%c",&c);     scanf("%d %d",&a,&b);     if(c == '+')       printf("%d\n",a+b);     else if(c == '-')       printf("%d\n",a-b);     else if(c == '*')       printf("%d\n",a*b);     else     {       if(a%b==0)         printf("%d\n",a/b);         else         printf("%.2f\n",(float)a/b);       }             }   system("pause");   return 0;     }