hdu6312 2018杭电多校第二场 1004 D Game 博弈

最新推荐文章于 2025-09-11 23:50:21 发布

转载最新推荐文章于 2025-09-11 23:50:21 发布 · 64 阅读

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原文链接：http://www.cnblogs.com/l609929321/p/9368581.html

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#java

本文介绍了一个简单的游戏胜负判定算法，游戏中两名玩家轮流从1到n的整数集合中选取一个数并删除该数及其所有倍数。文章给出了具体的实现代码，并通过分析证明了无论初始条件如何，先行玩家总能赢得游戏。

Game

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 184 Accepted Submission(s): 132

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

Source

2018 Multi-University Training Contest 2

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题意：给你一串数字，A和B分别可以对这串数字进行操作，每次操作可以删去一个数及他所有的因子，轮到谁时谁无法再进行操作谁就输了

首先考虑1-n，我们可以先不看1，则2-n一定有一个胜者，如果这个状态A胜的话，我们就可以选择那个让我们胜的数，因为1是所有数的因子所以1也会被删去，如果A输，我们可以先去掉1来转变状态。这样A还是会胜。

综述A一定会获胜

#include<iostream>
using namespace std;
int main() {
    int n;
    while( cin >> n ) {
        cout << "Yes" << endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/l609929321/p/9368581.html