Hduoj1170【水题】

/*Balloon Comes!
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 20159    Accepted Submission(s): 7592


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
*/
#include<stdio.h>
int main()
{
	char op;
	int i, j, k, m, n;
	float ans;
	scanf("%d", &n);
	while(n--)
	{
		getchar();
		scanf("%c%d%d", &op, &k, &m);
		if(op == '-')
		{
			ans = k - m;
		}
		else if(op == '+')
		{
			ans = k + m;
		}
		else if(op == '*')
		{
			ans = k * m;
		}
		else 
		{
			ans = (float)k / (float)m;
		}
		if(ans == (int) ans)
		printf("%d\n", (int)ans);
		else
		printf("%.2f\n", ans);
	}
	return 0;
}