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Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15008    Accepted Submission(s): 9158


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.

Author

LL

Source

​​校庆杯Warm Up​​


 

题意：

先把这些灯标上号，1 2 3 4 5 6 7 8 ……无穷 首先全是关的，也就是全是0 第一次操作 ，标号是1的倍数，全都变成相反的状态，也就是全变成1.。 第二次操作 ，标号是2的倍数，全都变成相反的状态，你可以看下，2 4 6……变成了0.。。 第三次操作 ，标号是3的倍数，全都变成相反的状态，你可以看下，3 6 9…… 他问你 N 号台灯最后 变成了 什么状态， 例如 1号灯，最后变成了1，不管多少次操作都是1.。 例如 5号灯 最后变成了0，不管多少次操作都是0.。 当操作次数大于N的时候 N的状态就不会改变了，因为N不会是M（M>N）的倍数。。

思路很简单就是求n有几个约数（包括1和自身）如果有奇数个约数，则是变奇数次，结果也就是1；否则为0

AC代码：



#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<algorithm>
typedef long long LL;
using namespace std;
int main()
{
    
    int n,k,i;
    while(scanf("%d",&n)!=EOF)
    {
        k=0;
        for(i=1;i<=n;i++)
        {
            if(n%i==0)
                k++;
        }
        if(k%2==0)
            printf("0\n");
        else
            printf("1\n");
    }
    return 0;
}