題意：給出n,m,p輸出C(n,m)%p p不一定是素數

#include <iostream>//ex_Lucas模板
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
//#define IO  ios::sync_with_stdio(false),cin.tie(0), cout.tie(0);
//#pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long ll;
void ex_gcd(ll a, ll b, ll &d, ll &x, ll &y)
{ if (!b) { x = 1; y = 0; d = a; } else { ex_gcd(b, a%b, d, y, x); y -= x * (a / b); }; }
int gcd(int a, int b) { return b ? gcd(b, a%b) : a; }
int lcm(int a,int b){return a/gcd(a,b)*b;}//Ïȳýºó³Ë·ÀÒç³ö
//int inv_exgcd(int a, int m) { int d, x, y;ex_gcd(a, m, d, x, y);return d == 1 ? (x + m) % m : -1; }
ll reverse(ll a, ll m) { ll d, x, y;ex_gcd(a, m, d, x, y);return d == 1 ? (x + m) % m : -1; }
typedef long long ll;
const int maxn=1e5;
using namespace std;
ll cnt=0;
int a[maxn];
int primer[maxn];
void isprime()
{
    cnt=0;
    for(int i=2;i<maxn;++i)
    {
        if(!a[i])
        {
            primer[cnt++]=i;
            for(int j=i*2;j<maxn;j+=i)
                a[j]=1;
        }
    }
}
ll Pow(ll a,ll n,ll mod)
{
    ll ans=1;
    while(n)
    {
        if(n&1)
            ans=ans*a%mod;
        a=a*a%mod;
        n>>=1;
    }
    return ans;
}
ll C(ll n,ll p,ll pk)
{
    if(n==0)return 1;
    ll ans=1;
    for(ll i=2;i<=pk;++i)
        if(i%p)ans=ans*i%pk;
    ans=Pow(ans,n/pk,pk);
    for(ll k=n%pk,i=2;i<=k;++i)
        if(i%p)ans=ans*i%pk;
    return ans*C(n/p,p,pk)%pk;
}
//ll reverse(ll a,ll m)
//{
//    if(!a)return 0;
//    ll y=0,x=1,r=a%m,q,t,M=m;
//    if(r<0)r+=m;
//    while(m%r)
//    {
//        a=m,m=r,q=a/m,r=a%m;
//        t=x,x=y-x*q,y=t;
//    }
//    if(r!=1)return 0;
//    if(x<0)x+=M;
//    return x;
//}
ll ex_lucas(ll n,ll m,ll p,ll pi,ll pk) //C(n,m)%p (p非素數) pi:p分解的一個素因子,pk：pi素因子對應的冪次
{
    ll i,j,k=0,a,b,c,ans;
    a=C(n,pi,pk),b=C(m,pi,pk),c=C(n-m,pi,pk);
    for(i=n;i;i/=pi)k+=i/pi;
    for(i=m;i;i/=pi)k-=i/pi;
    for(i=n-m;i;i/=pi)k-=i/pi;
    ans=a*reverse(b,pk)%pk*reverse(c,pk)%pk*Pow(pi,k,pk)%pk;
    return ans*(p/pk)%p*reverse(p/pk,pk)%p;
}
ll solve(ll n,ll m,ll x)
{
    ll ans=0,p=x;
    isprime();
    for(ll i=0;primer[i]*primer[i]<=x;++i)
    {
        if(x%primer[i]==0)
        {
           ll pk=1;
            while(x%primer[i]==0)
                x/=primer[i],pk*=primer[i];
            ans=(ans+ex_lucas(n,m,p,primer[i],pk))%p;
        }
    }
    if(x>1)
        ans=(ans+ex_lucas(n,m,p,x,x))%p;
    return  ans;
}
int main()
{
    ll n,m,p;
    cin>>n>>m>>p;
    ll ans=solve(n,m,p);
    printf("%lld\n",ans);
    return 0;
}