HDU 1170 Balloon Comes!

http://acm.hdu.edu.cn/showproblem.php?pid=1170

 

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12972 Accepted Submission(s): 4599

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

 

分析：虽然是水体不过 The result should be rounded to 2 decimal places If and only if it is not an integer.就是说如果是除法，并且结果为整数，就要输出整数。

代码如下：

#include<stdio.h>
int main()
{
  int T;
  char t;
  int a,b;
  scanf("%d%*c",&T);
  while(T--)
  {
      scanf("%c %d %d%*c",&t,&a,&b);
      if(t=='+')
            printf("%d\n",a+b);
            if(t=='-')
            printf("%d\n",a-b);
      if(t=='*')
            printf("%d\n",a*b);
      if(t=='/')
            {
              if(a%b==0)
          printf("%d\n",a/b);
          else
      printf("%.2lf\n",(double)a/b);
            }
  }
  return 0;
}