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

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 fif

th operation : 1 0 0 1 0 … The later operations cannot change the condition of the fifth lamp any more. So the answer is 0.

分析：

1.本题的意思是进行无穷次开关之后，输出第几盏灯的亮暗状态；只有在是i的倍数，才switch一次。偶数次的swtich是不变的0，奇数次才变化为1；

因此只需要计算给定n中因子的奇偶性就可以了，代码1可以AC。

2.再思考下，其实会发现乘积因子都是一对对的，如16=2*8=1*16=4*4 ，其实只要不是完全平方数，那因子总数一定是偶数，只有完全平方数才是奇数，这样只要判断是否完全平方数既可，很简单。

//1，

#include <iostream>
#include <stdlib.h>
using namespace std;
int main()
{
    int n,i;
    int count;
    while(scanf("%d",&n)!=EOF)
    {
        count = 0;
        for(i=1;i<=n;i++)
            if(n%i==0)
                count++;
        if(count & 0x01)
            printf("1
");
        else
            printf("0
");
        /*
        printf("%d
",count &0x01);//判断奇数为1
        */
    }
    return 0;
}

//判断完全平方数)

printf("%d
",((int)sqrt((double)n)*(int)sqrt((double)n))==n);