Switch Game + 枚举

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探讨一串初始关闭的灯在经历一系列操作后的最终状态。每次操作针对可被当前步骤数整除的灯，将其状态反转。通过预先计算并存储1到10^5范围内所有灯的最终状态来高效解决问题。

Switch Game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8468 Accepted Submission(s): 5017

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 ).

/*
题意：
一条线上有无数个灯，刚开始都是off状态，现在进行下列操作：第几下按开关，就将灯的编号能整除i的等置为相反的
状态，灯的编号为1.....inf。
思路：
类似于打表，由于n<=1000000，所以枚举100000次操作，在每次操作中将相关的灯置为相反的状态。
*/
#include<iostream>
using namespace std ;
#define MAX 100003
int a[MAX];
void fun()
{
	memset(a , 0 , sizeof(a)) ;
	for( int i = 1; i <= 100000 ; i ++)  //枚举100000个操作，并在每个操作中中将相关等状态置为相反
	{
		for( int j = i ; j <= 100000 ; j += i )
			if(a[j])
				a[j] = 0;
			else
				a[j] = 1;
	}
}
int main(void)
{
	fun();
	int n ;
	while(cin >>n )
	{
		cout << a[n] << endl;
	}
	return 0 ;
}

Sample Input