2019ICPC徐州网络赛 A.Who is better?——斐波那契博弈&&扩展中国剩余定理

题意

有一堆石子，两个顶尖聪明的人玩游戏，先取者可以取走任意多个，但不能全取完，以后每人取的石子数不能超过上个人的两倍。石子的个数是通过模方程组给出的。

题目链接

分析

斐波那契博弈有结论：当且仅当石子数为斐波那契数时，先手必败。

又因为 $n \leq 10^{15}$，在这个范围内的斐波那契数只有72个，可以预处理出来。

注意会爆long long !!

#include<iostream>
#include<cstdio>
#define LL __int128
using namespace std;

const LL MAXN = ;
LL fibo[] = {,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,};

int K;
LL C[MAXN], M[MAXN], x, y;
LL gcd(LL a, LL b) {
    return b ==  ? a : gcd(b, a % b);
}
LL exgcd(LL a, LL b, LL &x, LL &y) {
    if (b == ) {x = , y = ; return a;}
    LL r = exgcd(b, a % b, x, y), tmp;
    tmp = x; x = y; y = tmp - (a / b) * y;
    return r;
}
LL inv(LL a, LL b) {
    LL r = exgcd(a, b, x, y);
    while (x < ) x += b;
    return x;
}
int main() {
    scanf("%d", &K);
    for (int i = ; i <= K; i++)  //x = C[i](mod M[i])
    {
        long long _m, _c;
        scanf("%lld%lld", &_m, &_c);
        M[i] = _m, C[i] = _c;
    }
    bool flag = ;
    for (LL i = ; i <= K; i++) {
        LL M1 = M[i - ], M2 = M[i], C2 = C[i], C1 = C[i - ], T = gcd(M1, M2);
        if ((C2 - C1) % T != ) {flag = ; break;}
        M[i] = (M1 * M2) / T;  //可能爆long long
        C[i] = ( inv( M1 / T , M2 / T ) * (C2 - C1) / T ) % (M2 / T) * M1 + C1;
        C[i] = (C[i] % M[i] + M[i]) % M[i];
    }
    //printf("%lld\n", flag ? C[K] : -1);
    if(flag == )  printf("Tankernb!\n");
    else
    {
        bool flg = false;
        for(int i = ;i < ;i++)
            if(fibo[i] == C[K])
            {
                flg = true;
                break;
            }
        if(flg)  printf("Lbnb!\n");
        else printf("Zgxnb!\n");
    }

    return ;
}