HDU 3032 Nim or not Nim? Lasker's Nim

来源：互联网发布：123 网络编辑：程序博客网时间：2024/06/12 00:17

题目大意：

就是现在初始有多堆石子, 每次可以将任意一堆拿走任意个或者将这一对分成两堆(不为空), 拿走最后一颗石子的人胜利

问谁会获得胜利

大致思路：

这个是Lasker's Nim 的模型, 我是打表找出SG函数的规律做的...

SG[4*k + 1] = 4*k + 1;

SG[2*k + 2] = 4*k + 2;

SG[4*k + 3] = 4*k + 4;

SG[4*k + 4] = 4*k + 3;

然后利用游戏的和将每一堆石子作为子游戏, 得到异或和判断是否为0就行了

代码如下：

Result : Accepted Memory : 1628 KB Time : 0 ms

/* * Author: Gatevin * Created Time:  2015/5/2 9:16:05 * File Name: Rin_Tohsaka.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;#define foreach(e, x) for(__typeof(x.begin()) e = x.begin(); e != x.end(); ++e)#define SHOW_MEMORY(x) cout<<sizeof(x)/(1024*1024.)<<"MB"<<endl/* * SG[4*k + 1] = 4*k + 1; * SG[2*k + 2] = 4*k + 2; * SG[4*k + 3] = 4*k + 4; * SG[4*k + 4] = 4*k + 3; */int s[1000010];int main(){    int T;    scanf("%d", &T);    int n;    while(T--)    {        scanf("%d", &n);        for(int i = 0; i < n; i++)            scanf("%d", s + i);        for(int i = 0; i < n; i++)            switch(s[i] % 4)            {                case 0: s[i] = s[i] - 1; break;                case 1: s[i] = s[i]; break;                case 2: s[i] = s[i]; break;                case 3: s[i] = s[i] + 1; break;            }        int sg = 0;        for(int i = 0; i < n; i++)            sg ^= s[i];        if(sg) printf("Alice\n");        else printf("Bob\n");    }    return 0;}

0 0