HDU_2053

Problem Description

There are many lamps in a line. All of them are off at first. A series of operations are carried out on these lamps. On the i-th operation, the lamps whose numbers are the multiple of i change the condition ( on to off and off to on ).

 

Input

Each test case contains only a number n ( 0< n<= 10^5) in a line.

 

Output

Output the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).

 

Sample Input

1 5

 

Sample Output

1 0

Hint

hint

Consider the second test case: The initial condition : 0 0 0 0 0 … After the first operation : 1 1 1 1 1 … After the second operation : 1 0 1 0 1 … After the third operation : 1 0 0 0 1 … After the fourth operation : 1 0 0 1 1 … After the fifth operation : 1 0 0 1 0 … The later operations cannot change the condition of the fifth lamp any more. So the answer is 0.

#include <cstdio>
int main()
{
    int n, ans;
    while(~scanf("%d",&n))
        {
            ans=1;
            for(int i=1;i<=n;i++)
                {
                    if(n%i==0)
                        {
                            ans=-1*ans;    
                        }    
                }
            printf(ans==1?"0
":"1
");
        }
    return 0;    
}