【hdu 3032】Nim or not Nim?

最新推荐文章于 2021-01-27 14:01:31 发布

weixin_30835923

最新推荐文章于 2021-01-27 14:01:31 发布

阅读量47

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/AWCXV/p/7626735.html

本文介绍了一种基于经典Nim游戏的新变种，允许玩家分离堆叠，并提供了通过计算sg函数来确定胜者的策略。通过分析游戏规则和编写计算程序，得出了获胜条件的关键在于最终异或值是否为0。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2016 Accepted Submission(s): 1048

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

【题目链接】:http://acm.hdu.edu.cn/showproblem.php?pid=3032

【题解】

可以通过sg[i]=mex{sg[0..i-1],sg[x]^sg[y]}来计算所有的sg函数(部分)
(mex是不属于这个集合的最小整数,且其中x+y==i)
如
sg[0]=0;
sg[1]=1
sg[2] = mex(sg[0],sg[1],sg[1]^sg[1])=2
sg[3] = mex(sg[0],sg[1],sg[2],sg[1]^sg[2]) = 4
…
写一个打表的程序算一下,找下规律
->
sg[4n+1]=4n+1,sg[4n+2]=4n+2;
sg[4n+3]=4n+4;
sg[4n+4] = 4n+3;
n∈N
然后用组合博弈的解决办法求异或值;
为0则先手输,否则先手赢;
【打表程序↓(0..50的sg函数值)】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 100;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int sg[100];
bool flag[100];

int main()
{
    freopen("F:\\rush.txt","r",stdin);
    sg[0] = 0;sg[1] = 1;
    rep1(i,2,50)
    {
        memset(flag,0,sizeof flag);
        rep1(j,1,i/2)
            flag[sg[j]^sg[i-j]] = true;
        rep1(j,0,i-1)
            flag[sg[j]] = true;
        rep1(j,0,50)
            if (!flag[j])
            {
                sg[i] = j;
                break;
            }
    }
    rep1(i,0,50)
    {
        printf("sg[%d]=%d\n",i,sg[i]);
    }
    return 0;
}

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

//const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int main()
{
    /*
    sg[4n+1]=4n+1,sg[4n+2]=4n+2;
    sg[4n+3]=4n+4;
    sg[4n+4] = 4n+3;
    */
    //freopen("F:\\rush.txt","r",stdin);
    int T;
    rei(T);
    while (T--)
    {
        int n;
        LL judge = 0;
        rei(n);
        rep1(i,1,n)
        {
            LL x,temp,sg;
            rel(x);
            temp = x%4;
            if (temp==0)
                sg = ((x/4)-1)*4+3;
            if (temp==1 || temp==2)
                sg = x;
            if (temp ==3)
                sg = (x/4)*4+4;
            judge = judge ^ sg;
        }
        if (judge==0)
            puts("Bob");
        else
            puts("Alice");
    }
    return 0;
}