CF 2015 ICL, Finals, Div. 1 J. Ceizenpok’s formula [Lucas定理]

http://codeforces.com/gym/100633/problem/J


Lucas定理P不是质数裸题

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
inline ll read(){
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

ll n,m,MOD;
ll Pow(ll a,ll b,ll P){
    ll ans=1;
    for(;b;b>>=1,a=a*a%P)
        if(b&1) ans=ans*a%P;
    return ans;
}
void exgcd(ll a,ll b,ll &d,ll &x,ll &y){
    if(b==0) d=a,x=1,y=0;
    else exgcd(b,a%b,d,y,x),y-=(a/b)*x;
}
ll Inv(ll a,ll n){
    ll d,x,y;
    exgcd(a,n,d,x,y);
    return d==1?(x+n)%n:-1;
}
ll Fac(ll n,ll p,ll pr){
    if(n==0) return 1;
    ll re=1;
    for(ll i=2;i<=pr;i++) if(i%p) re=re*i%pr;
    re=Pow(re,n/pr,pr);
    ll r=n%pr;
    for(int i=2;i<=r;i++) if(i%p) re=re*i%pr;
    return re*Fac(n/p,p,pr)%pr;
}
ll C(ll n,ll m,ll p,ll pr){
    if(n<m) return 0;
    ll x=Fac(n,p,pr),y=Fac(m,p,pr),z=Fac(n-m,p,pr);
    ll c=0;
    for(ll i=n;i;i/=p) c+=i/p;
    for(ll i=m;i;i/=p) c-=i/p;
    for(ll i=n-m;i;i/=p) c-=i/p;
    ll a=x*Inv(y,pr)%pr*Inv(z,pr)%pr*Pow(p,c,pr)%pr;
    return a*(MOD/pr)%MOD*Inv(MOD/pr,pr)%MOD;
}

ll Lucas(ll n,ll m){
    ll x=MOD,re=0;
    for(ll i=2;i<=MOD;i++) if(x%i==0){
        ll pr=1;
        while(x%i==0) x/=i,pr*=i;
        re=(re+C(n,m,i,pr))%MOD;
    }
    return re;
}
int main(){
    //freopen("in","r",stdin);
    n=read();m=read();MOD=read();
    printf("%I64d",Lucas(n,m));
}
原文地址:https://www.cnblogs.com/candy99/p/6399577.html