HDU 1536 S-Nim (组合游戏+SG函数)

时间：2022-07-23 08:25:38

题意：针对Nim博弈，给定上一个集合，然后下面有 m 个询问，每个询问有 x 堆石子 ，问你每次只能从某一个堆中取出 y 个石子，并且这个 y 必须属于给定的集合，问你先手胜还是负。

析：一个很简单的博弈，对于每组数据，要先处理出SG函数， 然后使用组合游戏和来解决就ok了，对于求sg函数，很明显，就是求所有的mex，也就是未出现过的最小自然数。最后取异或就ok了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#include <numeric>

#define debug() puts("++++")

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define pu push_up

#define pd push_down

#define cl clear()

#define lowbit(x) -x&x

//#define all 1,n,1

#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 10000 + 10;

const int maxm = 100 + 2;

const LL mod = 100000000;

const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1};

const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

int a[maxm], g[maxn], cnt[maxm];

int main(){

  while(scanf("%d", &n) == 1 && n){

    for(int i = 0; i < n; ++i)  scanf("%d", a+i);

    sort(a, a + n);

    g[0] = 0;

    for(int i = 1; i <= 10000; ++i){

      for(int j = 0; j < n && i >= a[j]; ++j)

        cnt[g[i-a[j]]] = i;

      for(int j = 0; j <= n; ++j)

        if(cnt[j] != i){ g[i] = j;  break; }

    }

    scanf("%d", &m);

    while(m--){

      int x;  scanf("%d", &x);

      int ans = 0;

      while(x--){

        int y;  scanf("%d", &y);

        ans ^= g[y];

      }

      if(ans == 0)  putchar('L');

      else putchar('W');

    }

    putchar('\n');

  }

  return 0;

}

　　

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