斐波那契博弈 hdu2516 取石子游戏

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题意：1堆石子n个，第一个人可以取任意个数但不能全部取完，以后每次拿的个数不能超过上一次对手拿的个数的2倍，轮流拿石子，问先手是否必赢

思路：斐波那契博弈，后手赢的情况的数字会呈现斐波那契数列。

#include<map>
#include<set>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define fuck(x) cout<<"["<<x<<"]"
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout)
using namespace std;
typedef long long LL;

const int MX = 50;
LL F[MX];

void presolve() {
    F[1] = 1; F[2] = 2;
    for(int i = 3; i < 50; i++) {
        F[i] = F[i - 1] + F[i - 2];
    }
}

bool solve(int n) {
    for(int i = 1; i < MX; i++) {
        if(F[i] == n) return false;
    }
    return true;
}

int main() {
    presolve();
    int n; //FIN;
    while(~scanf("%d", &n), n) {
        printf("%s\n", solve(n) ? "First win" : "Second win");
    }
    return 0;
}