hdoj 1170 Balloon Comes!(水)

 

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25161    Accepted Submission(s): 9511

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
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    代码：
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    
#include<stdio.h>
int main()
{
    int t,a,b;
    char c;
    scanf("%d",&t);
    while(t--)
    {
    	getchar();//吸收空格 
        scanf("%c %d %d",&c,&a,&b);
       //或者 scanf("%c%d%d",&c,&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);//注意要求，另加强制转换 
        }
    }
    return 0;
 }