杭电 HDU 1170 Balloon Comes!

Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22017    Accepted Submission(s): 8291

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

 这个题 真他妈的傻逼 气死我啦 靠 今天本来就不爽

#include<stdio.h>

int main()
{int a,b;
	int T;
scanf("%d",&T);
	char ch[2];
	while(T--)
	{
		scanf("%s%d%d",ch,&a,&b);
		switch(ch[0])
		{
		case '+':
			printf("%d\n",a+b);
			break;
		case'-':
		printf("%d\n",a-b);
			break;
		case'*':
			printf("%d\n",a*b);
			break;
		case'/':
			if(a%b==0)
				printf("%d\n",a/b);
			else 
				printf("%.2f\n",a/(b*1.0));
			break;
		
		}
		
	}
	return 0;
}