Balloon Comes!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 23182 Accepted Submission(s): 8774



[align=left]Problem Description[/align]
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!

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

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

[align=left]Sample Output[/align]

3
-1
2
0.50

#include<stdio.h>
int main()
{
int  n,a,b;
char k;
scanf("%d",&n);
while(n--)
{
getchar();
scanf("%c %d%d",&k,&a,&b);
if(k=='+')
printf("%d\n",a+b);
if(k=='-')
printf("%d\n",a-b);
if(k=='*')
printf("%d\n",a*b);
if(k=='/')
{
if(a%b==0)
printf("%d\n",a/b);
else
printf("%.2f\n",a*1.0/b);
}
}
return 0;
}