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HDU 3032 Nim or not Nim? （sg函数求解）

2024-07-10 22:39:41   224人阅读

Nim or not Nim?


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
 


题目大意：

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。(1 ≤ N ≤ 10^6, 1 ≤ S[i] ≤ 2^31 - 1)


解题思路：

对于一个给定的有向无环图，定义关于图的每个顶点的Sprague-Grundy函数g如下：g(x)=mex{ g(y) | y是x的后继 },这里的g（x）即sg[x]
例如：取石子问题，有1堆n个的石子，每次只能取{1，3,4}个石子，先取完石子者胜利，那么各个数的SG值为多少？
sg[0]=0,
n=1时，可以取走{1}个石子，剩余{0}个，mex{sg[0]}={0}，故sg[1]=1;
n=2时，可以取走{1}个石子，剩余{1}个，mex{sg[1]}={1}，故sg[2]=0；
n=3时，可以取走{1,3}个石子，剩余{2,0}个，mex{sg[2],sg[0]}={0,0}，故sg[3]=1;
n=4时，可以取走{1,3,4}个石子，剩余{3,1,0}个，mex{sg[3],sg[1],sg[0]}={1,1,0},故sg[4]=2;
n=5时，可以取走{1,3,4}个石子，剩余{4,2,1}个，mex{sg[4],sg[2],sg[1]}={2,0,1},故sg[5]=3；
以此类推.....
     x  0  1  2  3  4  5  6  7  8....
sg[x] 0  1  0  1  2  3  2  0  1....

所以，对于这题：

sg[0]=0

sg[1]=mex{sg[0] }=1

sg[2]=mex{sg[0],sg[1],sg[1,1] }=mex{0,1,1^1}=2;

sg[3]=mex{sg[0],sg[1],sg[2],sg[1,2]}=mex{0,1,2,1^2}=mex{0,1,2,3}=4;

sg[4]=mex{sg[0],sg[1],sg[2],sg[3],sg[1,3],sg[2,2]}=mex{0,1,2,4,5,0}=3;

              ..............................................................................

可以发现：sg[4*k+1]=4*k+1,sg[4*k+2]=4*k+2,  sg[4*k+3]=4*k+4,sg[4*k+4]=4*k+3


解题代码：


#include <iostream>
using namespace std;

int main(){
    int t;
    cin>>t;
    while(t-- >0){
        int sg=0,n,x;
        cin>>n;
        for(int i=0;i<n;i++){
            cin>>x;
            if(x%4==0) sg^=x-1;
            else if(x%4==3) sg^=x+1;
            else sg^=x;
        }
        if(sg) cout<<"Alice"<<endl;
        else cout<<"Bob"<<endl;
    }
    return 0;
}




 


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