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 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 <iostream>
#include <cstdio>
using namespace std;
int main()
{
int n;
int cnt;
while (scanf("%d", &n) != EOF)
{
cnt = 0;
for (int i = 1; i <= n; i++)
if (0 == n % i)
cnt++;
printf("%d\n", cnt%2);
}
return 0;
}