HDU1170 Balloon Comes!

Balloon Comes!

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

Total Submission(s): 19874    Accepted Submission(s): 7486

[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

 

[align=left]Author[/align]
lcy
water

#include <stdio.h>

void f(char ch, int a, int b)
{
if(ch == '+') printf("%d\n", a + b);
else if(ch == '-') printf("%d\n", a - b);
else printf("%d\n", a * b);
}

int main()
{
int a, b, t;
char str[2];
scanf("%d", &t);
while(t--){
scanf("%s%d%d", str, &a, &b);
if(str[0] != '/') f(str[0], a, b);
else if(a % b == 0) printf("%d\n", a / b);
else printf("%.2lf\n", (double)a / b);
}
return 0;
}