HDU 1536 S-Nim (组合游戏+SG函数)

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;
}