1408122118-hd-Switch Game.cpp-优快云博客

本文讨论了在一系列操作下，特定位置的灯的状态变化规律。通过分析每个操作对灯的影响，最终确定了特定位置的灯在无限次操作后的状态。

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Switch Game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10697 Accepted Submission(s): 6521

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.

题目大意

开灯问题。有n盏灯，编号为1~n，第1个人把所有灯打开，第2个人按下所有编号为2 的倍数的开关（这些灯将被关掉），第3 个人按下所有编号为3的倍数的开关（其中关掉的灯将被打开，开着的灯将被关闭），依此类推。一共进行n此操作，问n此操作之后第n个灯的状态。

解题思路

因为只要求求第n个灯的状态，所以要简单很多。首先要建立一个数组，数组的下标即是灯的编号，每次操作看看是不是第n个灯的因子，是的话对其造成影响，其值加一，不是的话则无影响。最后对其值对2取余，因为其最开始为0，代表不亮，所以是奇数则亮（1），是偶数则灭（0）。

代码

#include<stdio.h>
#include<string.h>
int s[110000];
int main()
{
    int n;
    int i,j;
    while(scanf("%d",&n)!=EOF)
    {
        memset(s,0,sizeof(s));           //注意每组数据都要对数组进行初始化，避免上次操作影响这次的结果。
        for(i=1;i<=n;i++)
            if(n%i==0)
                s[n]++;
        printf("%d\n",s[n]%2);
    }
    return 0;
}