洛谷P2480 古代猪文

这道题把我坑了好久......

原因竟是CRT忘了取正数！

题意：求

指数太大了，首先用欧拉定理取模。

由于模数是质数所以不用加上phi(p)

然后发现phi(p)过大，不能lucas，但是它是个square free，可以分解质因数然后lucas然后CRT。

然后就没有然后了......模板套来套去......

注意CRT的结果可能是负数，要取正。

#include <cstdio>
#include <algorithm>
#define say(a) printf(#a); printf(" = %lld 
", a)

typedef long long LL;
const int N = 200010;
const LL MO = 999911659;
const LL prime[4] = {2, 3, 4679, 35617};

int T;
LL n, G;
LL nn[4][N], inv[4][N], invn[4][N], aa[4];

inline LL qpow(LL a, LL b, LL c) {
    LL ans = 1;
    while(b) {
        if(b & 1) {
            ans = ans * a % c;
        }
        a = a * a % c;
        b = b >> 1;
    }
    return ans;
}
LL C(int a, int b) {
    return nn[T][a] * invn[T][b] % prime[T] * invn[T][a - b] % prime[T];
}
LL lucas(int a, int b) {
    if(!b) {
        return 1;
    }
    return C(a % prime[T], b % prime[T]) * lucas(a / prime[T], b / prime[T]) % prime[T];
}
LL exgcd(LL a, LL b, LL &x, LL &y) {
    if(!b) {
        x = 1;
        y = 0;
        return a;
    }
    LL g = exgcd(b, a % b, x, y);
    std::swap(x, y);
    y -= (a / b) * x;
    return g;
}

inline void prework(int h) {
    inv[h][1] = 1;
    LL p = prime[h];
    for(int i = 2; i < p; i++) {
        inv[h][i] = (p - p / i) * inv[h][p % i] % p;
    }
    nn[h][0] = 1;
    invn[h][0] = 1;
    for(int i = 1; i < p; i++) {
        nn[h][i] = nn[h][i - 1] * i % p;
        invn[h][i] = invn[h][i - 1] * inv[h][i] % p;
    }
    return;
}

inline LL CRT() {

    LL M = MO - 1, ans = 0, b, t;
    for(int i = 0; i < 4; i++) {
        LL m = M / prime[i];
        exgcd(m, prime[i], t, b);
        t = (t % M + M) % M;
        ans += aa[i] * m % M * t % M;
        ans %= M;
    }

    return ans;
}

int main() {

    scanf("%lld%lld", &n, &G);
    if(G == MO) {
        printf("0");
        return 0;
    }
    for(int i = 0; i < 4; i++) {
        prework(i);
    }

    for(T = 0; T < 4; T++) {
        for(int i = 1; i * i <= n; i++) {
            if(n % i) {
                continue;
            }
            aa[T] += lucas(n, i);
            aa[T] %= prime[T];
            if(i * i < n) {
                aa[T] += lucas(n, n / i);
                aa[T] %= prime[T];
            }
        }
        //printf("aa[%d] = %lld 
", T, aa[T]);
    }

    LL ans = CRT();
    //printf("ans = %lld 
", ans);


    ans = qpow(G, ans, MO);
    printf("%lld
", ans);
    return 0;
}