HDU 3032 Nim or not Nim? （sg函数）

Nim or not Nim?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 623    Accepted Submission(s): 288

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

 

题意:Alice和Bob轮流取N堆石子，每堆S[i]个，Alice先，每一次可以从任意一堆中拿走任意个石子，也可以将一堆石子分为两个小堆。先拿完者获胜。

数据范围: (1 ≤ N ≤ 10^6, 1 ≤ S[i] ≤ 2^31 - 1).

思路: 此题为博弈中的—取走-分割游戏（这种游戏允许取走某些东西，然后将原来的一个游戏分成若干个相同的游戏）

由于数据范围,不能直接求sg值只能打表找规律;

有SJ 定理：

对于任意的一个 Anti-SG 游戏，如果我们规定当局面中所有单一游戏的 SG 值为 0 时游戏 结束，则先手必胜当且仅当以下两个条件满足任意一个：

(1)游戏的 SG 函数不为 0，且游戏中某个单一游戏的 SG 函数大于1。

(2)游戏的 SG 函数为 0，且游戏中没有单一游戏的 SG 函数大于 1。

Lasker's Nim游戏：每一轮允许两会中操作之一：①、从一堆石子中取走任意多个，②、将一堆数量不少于2的石子分成都不为空的两堆。

分析：很明显：sg(0) = 0，sg(1) = 1。

状态2的后继有：0，1和（1，1），他们的SG值分别为0，1，0，所以sg(2) =2。

状态3的后继有：0、1、2、（1，2），他们的SG值分别为0、1、2、3，所以sg(3) = 4。

状态4的后继有：0、1、2、3、（1，3）和（2，2），他们的SG值分别为0，1，2，4，5，0，所以sg(4) = 3.

再推一些，推测得到：对于所有的k >= 0，有 sg( 4k+1 ) = 4k+1； sg( 4k+2 ) = 4k+2； sg( 4k+3 ) = 4k+4； sg( 4k+4 ) = 4k+3。

假设游戏初始时有3堆，分别有2、5和7颗石子。三堆的SG函数值分别为2、5、8，他们的Nim和等于15.所以要走到P状态，就要使得第三堆的SG值变成7，可以将第三对按1和6分成两堆。

SG打表代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>

using namespace std;

const int N=1000010;

int sg[N];

int g(int x){
    int mex[1010];
    memset(mex,0,sizeof(mex));
    if(sg[x]!=-1)
        return sg[x];
    for(int i=x-1;i>=0;i--)
        mex[g(i)]=1;
    for(int i=1;i<=x/2;i++){
        int ans=0;
        ans^=g(i);
        ans^=g(x-i);
        mex[ans]=1;
    }
    for(int i=0;;i++)
        if(!mex[i])
            return sg[x]=i;
}

int main(){

    freopen("input.txt","r",stdin);

    int t,n;
    scanf("%d",&t);
    memset(sg,-1,sizeof(sg));
    while(t--){
        scanf("%d",&n);
        int x;
        for(int i=0;i<n;i++){
            scanf("%d",&x);
            g(x);
            printf("sg[%d]=%d
",x,sg[x]);
        }
        for(int i=0;i<=100;i++){
            printf("%d ",sg[i]);
            //if(i%10==0)
                //system("pause");
        }
        printf("
");
    }
    return 0;
}

可得规律：sg(4k)=4k-1;sg(4k+1)=4k+1;sg(4k+2)=4k+2;sg(4k+3)=4k+4;

本题AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>

using namespace std;

int main(){

    //freopen("input.txt","r",stdin);

    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        int ans=0,x;
        for(int i=0;i<n;i++){
            scanf("%d",&x);
            if(x%4==0)
                ans^=(x-1);
            else if(x%4==1 || x%4==2)
                ans^=x;
            else
                ans^=(x+1);
        }
        if(ans!=0)
            puts("Alice");
        else
            puts("Bob");
    }
    return 0;
}