hdu 1170 Balloon Comes!

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

 

#include <stdio.h>
#include <string.h>
const int maxn=500000;
int a[maxn+5];
int main()
{
    int t,a,b;
    char ch;
    scanf("%d",&t);
    while(t--)
    {
        getchar();
        scanf("%c %d%d",&ch,&a,&b);
        if(ch=='+')
            printf("%d
",a+b);
        else if(ch=='-')
            printf("%d
",a-b);
        else if(ch=='*')
            printf("%d
",a*b);
        else if(ch=='/')
        {
            if(a/b*b==a)
                printf("%d
",a/b);
            else
                printf("%.2lf
",(double)a/b);
        }
    }
    return 0;
}