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
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.

http://blog.csdn.net/hurmishine/article/details/51347080

 import java.util.Scanner;
import java.util.Arrays;

public class LampOnOff
{
    public static void main(String [] args)
    {
        Scanner input = new Scanner(System.in);

        System.out.print("输入灯的个数： ");
        int number = input.nextInt();

        int[] lamps = new int[number];
        Arrays.fill(lamps, 0);// 将所有的灯的状态初始化为 0

        for (int i = 1; i <= number; i++)
        {
            for (int j = 0; j < number; j++)
            {
                if ((j + 1) % i == 0)
                {
                    if (lamps[j] == 0)
                    {
                        lamps[j] = 1;
                    }
                    else if (lamps[j] == 1)
                    {
                        lamps[j] = 0;
                    }
                }
            }
        }

        System.out.println("第 " + number + " 的状态是： " + lamps[number - 1]);
    }
}