杭电 ACM 1.2.4

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Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2653 Accepted Submission(s): 869

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

3-120.50

Author

lcy

这题并不是很难，但是我第一次却没有A过去，而是得了个RA，原因很简单，是因为，当操作符为除号的时候，我没有做判断，到底是否可以除得尽，根据题目要求的是，只有结果不是整数的时候才要求结果精确到小数点后两位，郁闷，细节也很重要啊，提醒自己一下……切记，仔细读题

源代码：

#include<stdio.h>
int main(void)
{
   int n;
   int i;
   int a, b;
   char ch;
   scanf("%d", &n);
   getchar();
   for(i = 0;i < n;i++)
   {
       scanf("%c", &ch);
       getchar();
       scanf("%d", &a);
       getchar();
       scanf("%d", &b);
       getchar();
       switch(ch)
       {
       case '+':
           printf("%d\n", a + b);
           break;
       case '-':
           printf("%d\n", a - b);
           break;
       case '*':
           printf("%ld\n", (long)a * (long)b);
           break;
       case '/':
           if(a % b == 0)
               printf("%d\n", a / b);
           else
               printf("%0.2f\n", (double)a / (double)b);
           break;
       }
   }

   return 0;
}