HDU 1848 Fibonacci again and again SG函数做博弈

传送门

题意：

　　有三堆石子，双方轮流从某堆石子中去f个石子，知道不能去，问先手是否必胜，其中f为斐波那契数。

思路：

　　利用SG函数求解即可。

/*
* @Author: chenkexing
* @Date:   2019-01-13 16:17:46
* @Last Modified by:   chenkexing
* @Last Modified time: 2019-01-15 11:10:33
*/

#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 = 1009;
            ll f[50];
            int sg[maxn],s[maxn];

            void getSG(int n){

                  for(int i=1; i<=n; i++){
                        memset(s,0,sizeof(s));
                        for(int j=1; f[j] <= i && j<=20; j++){
                              s[sg[i-f[j]]] = 1;
                        }
                        for(int j=0; ; j++) if(!s[j]){
                              sg[i] = j;
                              break;
                        }
                  }
            }
int main(){ 
                  f[1] = 1;f[2] = 1;
                  for(int i=3; i<=20; i++) f[i] = f[i-1] + f[i-2];
                  getSG(1000);
                  int a,b,c;
                  while(~scanf("%d%d%d", &a, &b, &c) && a+b+c){
                        if(sg[a] ^ sg[b] ^ sg[c]) puts("Fibo");
                        else puts("Nacci");
                  }
                  return 0;
}

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