HDU_oj_2053 Switch Game

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

 分析：

题目大意为：有一列数，刚开始全为0，第一次将1的倍数的位置的数反转为1

第二次将2的倍数的位置的数反转为0

第三次将3的倍数的位置的数反转为1

……

……

注意点：

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
    double n;
    while(cin>>n)
    {
        n = sqrt(n);
        if(n == (int)n)
            cout<<1<<endl;
        else
            cout<<0<<endl;
    }
}