【hd水题】hdoj2053 Switch Game

最新推荐文章于 2019-02-25 20:49:41 发布

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最新推荐文章于 2019-02-25 20:49:41 发布

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分类专栏： ACM算法文章标签： hd水题

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本文探讨了在一系列操作下，灯泡从初始关闭状态变为开启状态的规律，通过数学分析得出结论，只需判断第n次操作时第n个灯泡被激活的次数是奇数还是偶数来确定其最终状态。

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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.

分析：只需要判断第n次第n个开了奇数次还是偶数次即可。开了奇数次，肯定是开着的状态。

代码如下：

#include<stdio.h> int main() {int i,j,n; while(scanf("%d",&n)!=EOF) {j=0; for(i=1;i<=n;i++) {if(n%i==0) j++; } if(j&1) printf("1\n"); else printf("0\n"); } return 0; }