Z

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

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.
Hint
hint

思路： 输入一个数n， 用一个计数函数观察它会被 改变多少次（ 它有几个约数，就会被改变几次） 然后 将它被改变的次数%2（两次一个周期） 这样就可以求得 答案。

#include<stdio.h>int main(){    int n,i,k;    while(~scanf("%d",&n))    {        k = 1;        for(i=1; i<n; i++)            if(n%i==0)                k ++;        printf("%d\n",k%2);    }    return 0;}