P4720 【模板】扩展卢卡斯定理/exLucas

题意：

$C_n^m\mathrm{m}\mathrm{o}\mathrm{d}p$
对于 100% 的数据，1≤m≤n≤10¹⁸，2≤p≤10⁶，不保证 p 是质数。

题解：

模板题，单纯写本文章记录板子

代码：

#include <cstdio>
using namespace std;typedef long long ll;ll mod;void exgcd(ll a, ll b, ll& x, ll& y)
{if (!b) {x= 1, y= 0;return;}exgcd(b, a % b, y, x);y-= a / b * x;return;
}inline ll inv(ll n, ll p)
{ll x, y;exgcd(n, p, x, y);return (x + p) % p;
}ll qpow(ll base, ll p, ll mod)
{ll ret= 1;for (; p; p>>= 1, base= base * base % mod)if (p & 1)ret= ret * base % mod;return ret;
}ll CRT(int n, ll* a, ll* m)
{ll M= 1, ret= 0;for (ll i= 1; i <= n; i++)M*= m[i];for (ll i= 1; i <= n; i++) {ll w= M / m[i];ret= (ret + a[i] * w % mod * inv(w, m[i]) % mod) % mod;}return (ret + mod) % mod;
}ll calc(ll n, ll q, ll qk)
{if (!n)return 1;ll ret= 1;for (ll i= 1; i <= qk; i++)if (i % q)ret= ret * i % qk;ret= qpow(ret, n / qk, qk);for (ll i= n / qk * qk + 1; i <= n; i++)if (i % q)ret= ret * (i % qk) % qk;return ret * calc(n / q, q, qk) % qk;
}ll multiLucas(ll n, ll m, ll q, ll qk)
{int cnt= 0;for (ll i= n; i; i/= q)cnt+= i / q;for (ll i= m; i; i/= q)cnt-= i / q;for (ll i= n - m; i; i/= q)cnt-= i / q;return qpow(q, cnt, qk) * calc(n, q, qk) % qk * inv(calc(m, q, qk), qk) % qk * inv(calc(n - m, q, qk), qk) % qk;
}ll exLucas(ll n, ll m, ll p)
{int cnt= 0;ll qk[20], a[20]; //存放所有的 q^k 和待合并答案的结果for (ll i= 2; i * i <= p; ++i) //质因数分解{if (p % i == 0) {qk[++cnt]= 1;while (p % i == 0)qk[cnt]*= i, p/= i;a[cnt]= multiLucas(n, m, i, qk[cnt]);}}if (p > 1)qk[++cnt]= p, a[cnt]= multiLucas(n, m, p, p);return CRT(cnt, a, qk); //CRT 合并答案
}int main()
{ll n, m, p;scanf("%lld %lld %lld", &n, &m, &p);mod= p;printf("%lld\n", exLucas(n, m, p));return 0;
}