组合博弈（NIM博弈和巴什博奕组合） hdu 1851

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本文介绍了一种简单的博弈游戏——石头堆博弈，并详细解析了其背后的算法原理。通过组合博弈论（NIM博弈和巴什博奕），文章阐述了如何判断后手玩家是否能获胜的策略，对于理解博弈论中的策略制定具有一定的指导意义。

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A Simple Game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 1178 Accepted Submission(s): 738

Problem Description

Agrael likes play a simple game with his friend Animal during the classes. In this Game there are n piles of stones numbered from 1 to n, the 1st pile has M ₁ stones, the 2nd pile has M ₂ stones, ... and the n-th pile contain M _n stones. Agrael and Animal take turns to move and in each move each of the players can take at most L ₁ stones from the 1st pile or take at most L ₂ stones from the 2nd pile or ... or take L _n stones from the n-th pile. The player who takes the last stone wins.

After Agrael and Animal have played the game for months, the teacher finally got angry and decided to punish them. But when he knows the rule of the game, he is so interested in this game that he asks Agrael to play the game with him and if Agrael wins, he won't be punished, can Agrael win the game if the teacher and Agrael both take the best move in their turn?

The teacher always moves first(-_-), and in each turn a player must takes at least 1 stones and they can't take stones from more than one piles.

Input

The first line contains the number of test cases. Each test cases begin with the number n (n ≤ 10), represent there are n piles. Then there are n lines follows, the i-th line contains two numbers M _i and L _i (20 ≥ M _i > 0, 20 ≥ L _i > 0).

Output

Your program output one line per case, if Agrael can win the game print "Yes", else print "No".

Sample Input

Sample Output

  
   Yes
No

Author

Agreal@TJU

Source

HDU 2007 Programming Contest - Final

//@auther yangZongJun
/********************************************//**
Date    : 2014/02/20
题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=1851
题    意：
        博弈
解题思路:组合博弈（NIM博弈和巴什博奕），从若干堆中给定可以取的最大的数，
判断后手胜输出yes
 ***********************************************/
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <string>
#include <cstring>

using namespace std;
#define PI acos(-1.0)
#define EPS 1e-8
const int MAXN = 100+5;
const int INF = 2100000000;

int cas, n, m, l;

int main()
{
//    freopen("input.txt", "r", stdin);
    while(~scanf("%d", &cas))
    {

        while(cas--)
        {
            int ans = 0;
            scanf("%d", &n);
            while(n--)
            {
                scanf("%d", &m);scanf("%d", &l);
                ans ^= (m%(l+1));
            }
            if(ans == 0) printf("Yes\n");
            else         printf("No\n");
        }

    }
    return 0;
}