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.

 1 #include <cstdio>

 2 int main()

 3 {

 4     int n, ans;

 5     while(~scanf("%d",&n))

 6         {

 7             ans=1;

 8             for(int i=1;i<=n;i++)

 9                 {

10                     if(n%i==0)

11                         {

12                             ans=-1*ans;    

13                         }    

14                 }

15             printf(ans==1?"0\n":"1\n");

16         }

17     return 0;    

18 }