HDU 3032 Nim or not Nim?（博弈，打表找规律）

最新推荐文章于 2025-09-12 18:08:49 发布

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最新推荐文章于 2025-09-12 18:08:49 发布

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文章标签： java

本文介绍了一种基于经典Nim博弈的变种游戏及其算法实现。在游戏中，玩家不仅可以移除堆中的物品，还可以将一个堆分成两个较小的堆。通过分析和计算策略函数（sg函数），文章给出了一种确定赢家的方法。

Nim or not Nim?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 540 Accepted Submission(s): 246

Problem Description

Nim is a two-player mathematic game of strategy in which players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap.

Nim is usually played as a misere game, in which the player to take the last object loses. Nim can also be played as a normal play game, which means that the person who makes the last move (i.e., who takes the last object) wins. This is called normal play because most games follow this convention, even though Nim usually does not.

Alice and Bob is tired of playing Nim under the standard rule, so they make a difference by also allowing the player to separate one of the heaps into two smaller ones. That is, each turn the player may either remove any number of objects from a heap or separate a heap into two smaller ones, and the one who takes the last object wins.

Input

Input contains multiple test cases. The first line is an integer 1 ≤ T ≤ 100, the number of test cases. Each case begins with an integer N, indicating the number of the heaps, the next line contains N integers s[0], s[1], ...., s[N-1], representing heaps with s[0], s[1], ..., s[N-1] objects respectively.(1 ≤ N ≤ 10^6, 1 ≤ S[i] ≤ 2^31 - 1)

Output

For each test case, output a line which contains either "Alice" or "Bob", which is the winner of this game. Alice will play first. You may asume they never make mistakes.

Sample Input

2 3 2 2 3 2 3 3

Sample Output

Alice Bob

Source

2009 Multi-University Training Contest 13 - Host by HIT

Recommend

gaojie

这题就是属于Nim博弈的类型。

可以先看论文了解。

其实就是先求得sg函数，然后n个数的sg值的异或和，为0，则败，非零则赢。

求sg值时打表了100内的，发现大部分是sg[i]=i.

除了i%4==0 则sg[i]=i-1;

i%4==3则sg[i]=i+1;

其余sg[i]=i;

//============================================================================
// Name        : HDU.cpp
// Author      : 
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
using namespace std;
/*
const int MAXN=1010;
int sg[MAXN];
bool vis[MAXN];

int mex(int v)
{
    if(sg[v]!=-1)return sg[v];
    memset(vis,false,sizeof(vis));
    int tmp;
    tmp=mex(0);
    vis[tmp]=true;
    for(int i=1;i<v;i++)
    {
        tmp=mex(v-i);
        vis[tmp]=true;
        tmp=mex(i)^mex(v-i);
        vis[tmp]=true;
    }
    for(int i=0;i<MAXN;i++)
        if(vis[i]==false)
        {
            sg[v]=i;
            break;
        }
    return sg[v];
}
*/
int mex(int v)
{
    if(v==0)return 0;
    if(v%4==0)return v-1;
    if(v%4==3)return v+1;
    return  v;

}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    //memset(sg,-1,sizeof(sg));
    //sg[0]=0;
    int T;
    int n;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int ans=0;
        int v;
        while(n--)
        {
            scanf("%d",&v);
            ans^=mex(v);
        }
        if(ans==0)printf("Bob\n");
        else printf("Alice\n");
    }
    return 0;
}