BZOJ 1951: [Sdoi2010]古代猪文 ExCRT+欧拉定理+Lucas

欧拉定理不要忘记！！

#include <bits/stdc++.h>
#define N 100000
#define ll long long
#define ull unsigned long long
#define setIO(s) freopen(s".in","r",stdin)
using namespace std;
int array[10]={2,3,4679,35617};
ll mult(ll x,ll y,ll mod)
{
    ll tmp=(long double)x/mod*y;
    return ((ull)x*y-tmp*mod+mod)%mod;
}
ll qpow(ll base,ll k,ll mod)
{
    ll tmp=1;
    for(;k;k>>=1,base=mult(base,base,mod)) if(k&1) tmp=mult(tmp,base,mod);
    return tmp;
}
struct Lucas
{
    int mod;
    int fac[N];
    ll inv(ll x)
    {
        return qpow(fac[x],mod-2,mod);
    }
    void init(int p)
    {
        mod=p;
        fac[0]=1;
        for(int i=1;i<=mod;++i) fac[i]=(ll)fac[i-1]*i%mod; // inv[i]=qpow(fac[i],mod-2,mod);
    }
    ll C(int n,int m)
    {
        if(m==0) return 1;
        if(n<m)  return 0;
        return (ll)(1ll*fac[n]*inv(m)%mod*inv(n-m)%mod)%mod;
    }
    ll solve(int n,int m)
    {
        if(m==0) return 1;
        return (solve(n/mod,m/mod)*C(n%mod,m%mod))%mod;
    }
}comb;
struct CRT
{
    int n;
    ll arr[N],brr[N];
    ll exgcd(ll a,ll b,ll &x,ll &y)
    {
        if(!b)
        {
            x=1,y=0;
            return a;
        }
        ll gcd=exgcd(b,a%b,x,y),tmp=x;
        x=y,y=tmp-a/b*y;
        return gcd;
    }
    ll ExCRT()
    {
        ll ans=arr[1],M=brr[1];
        for(int i=2;i<=n;++i)
        {
            ll a=M,b=brr[i],c=arr[i]-ans,gcd,x,y;
            gcd=exgcd(a,b,x,y),b=abs(b/gcd);
            x=x*(c/gcd),x=(x%b+b)%b;
            ans+=M*x;
            M*=brr[i]/__gcd(brr[i],M);
            ans=(ans%M+M)%M;
        }
        return ans;
    }
}crt;
int main()
{
    // setIO("input");
    int i,j;
    ll nn,gg,re,tmp,h;
    scanf("%lld%lld",&nn,&gg);
    if(gg%(999911659)==0)
    {
        printf("0\n");
        return 0;
    }
    for(i=0;i<4;++i)
    {
        comb.init(array[i]),re=0,h=nn;
        for(j=1;j*j<nn;++j)
        {
            if(nn%j==0)
            {
                re=(re+comb.solve((int)nn,j))%array[i];
                re=(re+comb.solve((int)nn,nn/j))%array[i];
            }
        }
        if(j*j==nn) re=(re+comb.solve((int)nn,j))%array[i];
        crt.arr[i+1]=re;
        crt.brr[i+1]=(ll)array[i];
    }
    crt.n=4;
    tmp=crt.ExCRT();
    printf("%lld\n",qpow(gg,tmp,999911659));
    return 0;
}