A Simple Game

本文介绍了一个简单的博弈游戏,玩家轮流从一堆或多堆石子中取石子,每次取石子的数量有一定限制。文章通过示例解释了如何使用尼姆博弈策略来确定先手玩家是否能赢得游戏。

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A Simple Game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 75 Accepted Submission(s): 52
 
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 M1 stones, the 2nd pile has M2 stones, ... and the n-th pile contain Mn stones. Agrael and Animal take turns to move and in each move each of the players can take at most L1 stones from the 1st pile or take at most L2 stones from the 2nd pile or ... or take Ln 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 Mi and Li (20 ≥ Mi > 0, 20 ≥ Li > 0).
 
Output
Your program output one line per case, if Agrael can win the game print "Yes", else print "No".
 
Sample Input
2
1
5 4
2
1 1
2 2
 
Sample Output
Yes
No
 
Author
Agreal@TJU
 
Source
HDU 2007 Programming Contest - Final
 
Recommend
lcy
 
/*
题意:有m堆石子,每堆石子有Mi个,每次最多取Li个,谁取到最后一个谁输

初步思路:对于一堆来说是八神博弈,对于整体来说是尼姆博弈
*/
#include<bits/stdc++.h>
using namespace std;
int t,n,m,l;
int res=0;
int main(){
    // freopen("in.txt","r",stdin);
    scanf("%d",&t);
    while(t--){
        res=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%d%d",&m,&l);
            res^=(m%(l+1));
        }
        printf(res?"No\n":"Yes\n");
    }
    return 0;
}

 

转载于:https://www.cnblogs.com/wuwangchuxin0924/p/6376487.html

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