HDOJ2053_Switch Game

这个题目就有意思了，要求输入一个数n，然后有一行数，都是0，我们可以设定为n个，然后是经过n次 的转换，转换有n次，假设当前使i，如果n%i==0的话，就变换一次（0变成1，1变成0）

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

15

 

Sample Output

10
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.
import java.util.Scanner;public class P2053 {private static Scanner scanner;public static void main(String[] args) {scanner = new Scanner(System.in);while(scanner.hasNext()){int n = scanner.nextInt();int k = 0;for (int i = 1; i <= n; i++) {if(n%i == 0){//如果是倍数就改变if(k==0){k = 1;}else {k = 0;}}}System.out.println(k);}}}