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

题意：有n盏灯，编号从1~n，进行n次操作，在进行i次操作的时候调整i的倍数的灯的编号（将关变成开，将开变成关），刚开始所有的灯都是关闭状态，问第n盏灯最后是什么状态。

解题思路：若n的约数的个数是奇数，那么是打开状态（即输出1），如果是偶数，输出 0，是关闭状态。

代码：

#include <stdio.h>

int main()
{
    int n,i,sum;
    while(scanf("%d",&n)!=EOF)
    {
        sum=0;
        for(i=1;i<=n;i++)
        {
            if(n%i==0)
                sum++;
        }
        printf("%d\n",sum%2);
    }
    return 0;
}