gym/102059/problem/I. Game on Plane SG函数做博弈

传送门：

题意：

　　给定一个正n边形的点。双方轮流连点成线，要求所画的线不能与之前的线相交。当某个人连成一个回路，这个人就输了。问先手必胜还是后手必胜。

思路：

　　SG函数，因为一条线相当于把图劈成了两半，所以每次用异或运算推过来。

/*
* @Author: chenkexing
* @Date:   2019-01-13 16:17:46
* @Last Modified by:   chenkexing
* @Last Modified time: 2019-01-15 18:33:24
*/

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue

typedef long long ll;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl ‘\n‘

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 1e9+7;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;

template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<‘0‘||ch>‘9‘) f|=(ch==‘-‘),ch=getchar();
    while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
    return x=f?-x:x;
}
/*-----------------------showtime----------------------*/
                  const int maxn = 5009;
                  int sg[maxn],s[maxn];
                  void getsg(int n){

                        for(int i=1; i<=n; i++){
                              memset(s, 0, sizeof(s));
                             for(int j=0; j<=i-2; j++){
                                    s[(sg[j] ^ sg[i-j-2])] = 1;
                             }
                              for(int j=0; ; j++){
                                    if(!s[j]) {
                                          sg[i] = j;
                                          break;
                                    }
                              }
                        }
                  }
int main(){
                  int T;
                  getsg(5000);
                  scanf("%d", &T);
                  while(T--){
                        int n;      scanf("%d", &n);
                        if(sg[n])puts("First");
                        else puts("Second");
                  }

                  return 0;
}

原文地址：https://www.cnblogs.com/ckxkexing/p/10274037.html