模板：exlucas

求$C_n^m mod p$，其中p不是质数且不保证p能分解为几个不同质数的乘积（也就是不能用crt合并）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define re register
#define int long long
using namespace std;
int n,m,p;
int q_pow(int a,int b,int p){
	int res=1;
	while(b){
		if(b&1) res=res*a%p;
		a=a*a%p;
		b>>=1;
	}
	return res;
}
inline void exgcd(re int a,re int b,re int &x,re int &y){
    if(!b){x=1,y=0;return;}
    exgcd(b,a%b,x,y);
    re int t=x;
    x=y,y=t-(a/b)*y;
}
inline int inv(re int a,re int b){
    re int x,y;
	exgcd(a,b,x,y);
    return (x%b+b)%b;
}
inline int crt(re int x,re int p,re int mod){
    return inv(p/mod,mod)*(p/mod)*x;
}
inline int fac(re int x,re int a,re int b){
    if(!x) return 1;
    re int ans=1;
    for(re int i=1;i<=b;i++)
    if(i%a) ans*=i,ans%=b;
    ans=q_pow(ans,x/b,b);
    for(re int i=1;i<=x%b;i++){
        if(i%a) ans*=i,ans%=b;
	}
    return ans*fac(x/a,a,b)%b;
}
inline int C(re int n,re int m,re int a,re int b){
    re int N=fac(n,a,b),M=fac(m,a,b),Z=fac(n-m,a,b),sum=0;
    for(re int i=n;i;i=i/a) sum+=i/a;
    for(re int i=m;i;i=i/a) sum-=i/a;
    for(re int i=n-m;i;i=i/a) sum-=i/a;
    return N*q_pow(a,sum,b)%b*inv(M,b)%b*inv(Z,b)%b;
}
inline int exlucas(re int n,re int m,re int p){
    re int t=p,ans=0;
    for(re int i=2;i*i<=p;i++){
        re int k=1;
        while(t%i==0)
            k*=i,t/=i;
        ans+=crt(C(n,m,i,k),p,k),ans%=p;
    }
    if(t>1) ans+=crt(C(n,m,t,t),p,t),ans%=p;
    return ans%p;
}
signed main(){
	scanf("%lld%lld%lld",&n,&m,&p);
	printf("%lld\n",exlucas(n,m,p));
	return 0;
}

质因数分解求组合数：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 1000006
#define re register
#define int long long
using namespace std;
int n,m,p;
int min(int a,int b){
	return a<b?a:b;
}
int max(int a,int b){
	return a>b?a:b;
}
inline int q_pow(re int a,re int b,re int p){
    int res=1;
    while(b){
        if(b&1) res=res*a%p;
        a=a*a%p;
        b>>=1;
    }
    return res%p;
}
int prime[MAXN],pr[MAXN],tot=0;
bool vis[MAXN];
void get_prime(int N){
    vis[1]=1;
    for(int i=2;i<=N;i++){
        if(!vis[i]) prime[++tot]=i,pr[i]=tot;
        for(int j=1;j<=tot&&i*prime[j]<=N;j++){
            vis[i*prime[j]]=1,pr[i*prime[j]]=j;
            if(!(i%prime[j])) break;
        }
    }
}
int d[MAXN];
void add(int x,int val){
	while(x!=1){
		d[pr[x]]+=val;
		x/=prime[pr[x]];
	}
}
inline int C(int n,int m,int p){
	int res=1;
	for(int i=1;i<=m;++i)
		add(n-i+1,1),add(i,-1);
    for(re int j=1;j<=tot;j++){
		for(re int k=1;k<=d[j];k++)
			(res*=prime[j])%=p;
	}
	return res;
}
signed main(){
	scanf("%lld%lld%lld",&n,&m,&p);
	get_prime(n);
	printf("%lld\n",C(n,m,p));
	return 0;
}