luogu P2480 [SDOI2010]古代猪文

https://www.luogu.com.cn/problem/P2480

读完题后不难得到要求的实际上是

\[g^{\sum_{d|n}\binom{n}{d}} \mod p \]

难点在求上面那个东西\(\mod (p-1)\)

直接\(Lucas\)显然会寄，把\(p-1\)质因数分解一下可以得到

\(999911658=2\times3\times4679\times35617\)

于是乎对于每个质数分别跑一遍，最后再用中国剩余定理合并起来即可

code:

#include<bits/stdc++.h>
#define ll long long 
#define N 2000050
using namespace std;
const int mod = 999911659;
int b[7] = {0, 2, 3, 4679, 35617};

ll qpow(ll x, ll y, int p) {
    ll ret = 1;
    for(; y; y >>= 1, x = x * x % p) if(y & 1) ret = ret * x % p;
    return ret;
}
ll fac[N], ifac[N];
void init(int n, int p) { n --;
    fac[0] = 1;
    for(int i = 1; i <= n; i ++) fac[i] = fac[i - 1] * i % p;
    ifac[n] = qpow(fac[n], p - 2, p);
    for(int i = n - 1; i >= 0; i --) ifac[i] = ifac[i + 1] * (i + 1) % p;
}
ll C(int n, int m, int p) {
    if(n < m) return 0;
    return fac[n] * ifac[m] % p * ifac[n - m] % p;
}
ll Lucas(int n, int m, int p) {
    if(!m) return 1;
    return Lucas(n / p, m / p, p) % p * C(n % p, m % p, p) % p;
}

ll ans = 0, a[N];
void crt() {
    int M = mod - 1;
    for(int i = 1; i <= 4; i ++) {
        int mi = M / b[i];
        (ans += 1ll * a[i] * mi % M * qpow(mi, b[i] - 2, b[i]) % M) %= M;
    }
}
int n, g;
int main() {
    scanf("%d%d", &n, &g);
    if(g % mod == 0) {
        printf("0");
        return 0;
    }

    for(int i = 1; i <= 4; i ++) {
        init(b[i], b[i]);

        for(int j = 1; j * j <= n; j ++) if(n % j == 0) {
            a[i] = (a[i] + Lucas(n, j, b[i])) % b[i];
         //   printf("%d  %d   %d   %lld\n", n, j, b[i], Lucas(n, j, b[i]));
            if(j * j != n) a[i] = (a[i] + Lucas(n, n / j, b[i])) % b[i];
        }
    }

    //for(int i = 1; i <= 4; i ++) printf("%lld %d\n", a[i], b[i]);
    crt();
    //printf("%lld\n", ans);
    printf("%lld", qpow(g, ans, mod));
    return 0;
}