HDU 6266 Hakase and Nano 【博弈论】

博弈论游戏策略解析

最新推荐文章于 2019-03-08 14:17:00 发布

转载最新推荐文章于 2019-03-08 14:17:00 发布 · 172 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/SCaryon/p/9089525.html

本文解析了一种特殊的Nim游戏变体，其中一方玩家在每轮可以连续取两次石子，探讨了不同条件下游戏胜负的策略。通过分析石子堆的数量和分布，总结出了简洁的胜负判断规则。

题面：

Hakase and Nano are playing an ancient pebble game (pebble is a kind of rock). There are n packs
of pebbles, and the i-th pack contains ai pebbles. They take turns to pick up pebbles. In each turn,
they can choose a pack arbitrarily and pick up at least one pebble in this pack. The person who
takes the last pebble wins.
This time, Hakase cheats. In each turn, she must pick pebbles following the rules twice continuously.
Suppose both players play optimally, can you tell whether Hakase will win?
Input
The first line contains an integer $T (1 <= T <= 20) $ representing the number of test cases.
For each test case, the first line of description contains two integers $n(1 <= n <= 106) and d (d = 1 or d = 2)$. If d = 1, Hakase takes first and if d = 2, Nano takes first. n represents the number of pebble
packs.
The second line contains n integers, the i-th integer ai (1 ai 109) represents the number of
pebbles in the i-th pebble pack.
Output
For each test case, print “Yes” or “No” in one line. If Hakase can win, print “Yes”, otherwise, print“No”.

Example

standard input	standard output
2
3 1
1 1 2	Yes
3 2
1 1 2	No

题目大意：

有两个人从N个石子堆中拿石子，规则类似Nim游戏。但不一样的是第一个人可以拿两次，第二个人只能拿一次。问最后谁能把石子都拿完。

大致思路：

面对博弈题我的思路是先把简单情况推一遍，然后分析。

如图，图中A,B表示谁先手，W,L表示在这种情况下，A是否能赢。

所以就有以下规律：

1.当 $n$ 为 3的倍数时，若每个石子堆都是1，A先手必输。只有一个数量大于1的石子堆，则B先手A必输

2.当 $n$ 为3的倍数加一时。若数量大于1的石子堆数量小于等于一。则B先手A必输。

其他情况均为A必胜。

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+7;
int main()
{
    ios::sync_with_stdio(false);
    bool flag=true;
    int t,n,d,a;
    cin>>t;
    while(t--)
    {
        flag=true;
        int cnt=0;
        cin>>n>>d;
        for(int i=0;i<n;++i){
            cin>>a;
            if(a>=2)
                cnt++;
        }
        int x=n%3;
        if(x==0){
            if(cnt==0 && d==1)
                flag=false;
            if(cnt==1 && d==2)
                flag=false;
        }else if(x==1){
            if(cnt<=1 && d==2)
                flag=false;
        }
        if(flag)
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
    return 0;
}

转载于:https://www.cnblogs.com/SCaryon/p/9089525.html