Switch Game

2019-11-06 06:27:49

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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 ). InputEach test case contains only a number n ( 0< n<= 10^5) in a line. OutputOutput the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).Sample Input

15Sample Output10
----------------------------我是滑稽的分割线------------------------------------------------------------------------
思路分析：
每个人进入都会改变灯的状态，第n个进入就会改变n的倍数的号的灯，比如第一个进所有数字都被改变，第二个进，2、4、6、8....状态会反转。
实质上就是求：一个数字的因子个数。
回顾总结：
该题暂未遇到坑。
#include <iostream>#include<string.h>/* run this PRogram using the console pauser or add your own getch, system("pause") or input loop */using namespace std;int lamp[100001];//求能被这个数整除的个数 int main(int argc, char** argv) {int n;while(cin>>n){	int ct=0;	for(int i=1;i<=n;i++)	{		if(n%i==0)		ct++;	 } 	 if(ct%2==0)	 {	 printf("0/n");	 }	 else	 {	  printf("1/n");	 }}	return 0;}