组合数取模（Lucas）

组合数取模（Lucas）

n，m较大且p不为素数的时候

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
ll pow(ll a, ll b, ll m)
{
    ll ans = 1;a %= m;while(b){
    if(b & 1)ans = (ans % m) * (a % m) % m;b /= 2;a = (a % m) * (a % m) % m;}ans %= m;return ans;
}
ll extgcd(ll a, ll b, ll& x, ll& y)
//求解ax+by=gcd(a, b)
//返回值为gcd(a, b)
{
    ll d = a;if(b){
    d = extgcd(b, a % b, y, x);y -= (a / b) * x;}else x = 1, y = 0;return d;
}
ll mod_inverse(ll a, ll m)
//求解a关于模上m的逆元
//返回-1表示逆元不存在
{
    ll x, y;ll d = extgcd(a, m, x, y);return d == 1 ? (m + x % m) % m : -1;
}ll Mul(ll n, ll pi, ll pk)//计算n! mod pk的部分值 pk为pi的ki次方
//算出的答案不包括pi的幂的那一部分
{
    if(!n)return 1;ll ans = 1;if(n / pk){
    for(ll i = 2; i <= pk; i++) //求出循环节乘积if(i % pi)ans = ans * i % pk;ans = pow(ans, n / pk, pk); //循环节次数为n / pk}for(ll i = 2; i <= n % pk; i++)if(i % pi)ans = ans * i % pk;return ans * Mul(n / pi, pi, pk) % pk;//递归求解
}ll C(ll n, ll m, ll p, ll pi, ll pk)//计算组合数C(n, m) mod pk的值 pk为pi的ki次方
{
    if(m > n)return 0;ll a = Mul(n, pi, pk), b = Mul(m, pi, pk), c = Mul(n - m, pi, pk);ll k = 0, ans;//k为pi的幂值for(ll i = n; i; i /= pi)k += i / pi;for(ll i = m; i; i /= pi)k -= i / pi;for(ll i = n - m; i; i /= pi)k -= i / pi;ans = a * mod_inverse(b, pk) % pk * mod_inverse(c, pk) % pk * pow(pi, k, pk) % pk;//ans就是n! mod pk的值ans = ans * (p / pk) % p * mod_inverse(p / pk, pk) % p;//此时用剩余定理合并解return ans;
}ll Lucas(ll n, ll m, ll p)
{
    ll x = p;ll ans = 0;for(ll i = 2; i <= p; i++){
    if(x % i == 0){
    ll pk = 1;while(x % i == 0)pk *= i, x /= i;ans = (ans + C(n, m, p, i, pk)) % p;}}return ans;
}int main()
{
    ll n, m, p;while(cin >> n >> m >> p){
    cout<<Lucas(n, m, p)<<endl;}return 0;
}