HDU6312 Game (多校第二场1004) 简单博弈

最新推荐文章于 2020-12-09 12:37:38 发布

weixin_33717117

最新推荐文章于 2020-12-09 12:37:38 发布

阅读量105

点赞数

CC 4.0 BY-SA版权

文章标签： java

原文链接：http://www.cnblogs.com/ACMerszl/p/9665370.html

本文介绍了一个简单的游戏胜负判断算法，游戏中两名玩家轮流从1到n的正整数集合中选择并移除其所有因数。文章通过转化问题，简化了胜负判断的过程，并提供了一个高效的解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1340 Accepted Submission(s): 891

Problem Description

Alice and Bob are playing a game.
The game is played on a set of positive integers from 1 to n.
In one step, the player can choose a positive integer from the set, and erase all of its divisors from the set. If a divisor doesn't exist it will be ignored.
Alice and Bob choose in turn, the one who cannot choose (current set is empty) loses.
Alice goes first, she wanna know whether she can win. Please judge by outputing 'Yes' or 'No'.

Input

There might be multiple test cases, no more than 10. You need to read till the end of input.
For each test case, a line containing an integer n. (

Output

A line for each test case, 'Yes' or 'No'.

Sample Input

Sample Output

Yes

之前邀请赛的原题，当是写了几个数发现的规律。但是不知道为什么。。。。

其实可以把 1~n 转化为 2~n

如果2~n 先手必败的话，那么先手可以第一次选1，把必败状态转移给后手；

如果2~n 先手必胜的话，多一个1其实是没有影响的。

证毕；

1 #include <bits/stdc++.h>
2 #define lowbit(x) (x)&(-x)
3 using namespace std;
4 int main()
5 {
6     int n;
7     while(~scanf("%d",&n))
8     cout<<"Yes"<<endl;
9 }