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 itis to see balloons floating around. You, one of the bestprogrammers in HDU, can get a very beautiful balloon if only youhave 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 rightnow!
Good Luck!

 

Input

Input contains multiple test cases. Thefirst line of the input is a single integer T(0<T<1000) which is the number oftest cases. T test cases follow. Each test case contains a char C(+,-,*, /) and two integers A andB(0<A,B<10000).Of course, we all knowthat A and B are operands and C is an operator.

 

Output

For each case, print the operationresult. The result should be rounded to 2 decimal places If andonly if it is not an integer.

 

Sample Input

4 + 1 2 - 12 * 1 2 / 1 2

 

Sample Output

3 -1 20.50

 

Author

lcy

 

分析：虽然是水体不过 The result should be roundedto 2 decimal places If and only if it is not aninteger.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;
}