【Luogu】P3927 SAC E#1 - 一道中档题 Factorial

【题目】洛谷10月月赛R1 提高组

【题意】求n!在k进制下末尾0的个数，n<=1e18,k<=1e16。

【题解】考虑10进制末尾0要考虑2和5，推广到k进制则将k分解质因数。

每个质因数在n!中的数量，以2为例是n/2+n/4+n/8...这样统计。（含x个就被统计x次）

最后得到凑出的k的个数就可以得到末尾0的个数。

分解质因数复杂度O(√k)，也使用pollard rho算法可以加速。

#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <stdio.h>

const int Times = ;
const int N = ;

using namespace std;
typedef long long LL;

LL ct, cnt;
LL fac[N], num[N];

LL gcd(LL a, LL b)
{
    return b? gcd(b, a % b) : a;
}

LL multi(LL a, LL b, LL m)
{
    LL ans = ;
    a %= m;
    while(b)
    {
        if(b & )
        {
            ans = (ans + a) % m;
            b--;
        }
        b >>= ;
        a = (a + a) % m;
    }
    return ans;
}

LL quick_mod(LL a, LL b, LL m)
{
    LL ans = ;
    a %= m;
    while(b)
    {
        if(b & )
        {
            ans = multi(ans, a, m);
            b--;
        }
        b >>= ;
        a = multi(a, a, m);
    }
    return ans;
}

bool Miller_Rabin(LL n)
{
    if(n == ) return true;
    if(n <  || !(n & )) return false;
    LL m = n - ;
    int k = ;
    while((m & ) == )
    {
        k++;
        m >>= ;
    }
    for(int i=; i<Times; i++)
    {
        LL a = rand() % (n - ) + ;
        LL x = quick_mod(a, m, n);
        LL y = ;
        for(int j=; j<k; j++)
        {
            y = multi(x, x, n);
            if(y ==  && x !=  && x != n - ) return false;
            x = y;
        }
        if(y != ) return false;
    }
    return true;
}

LL pollard_rho(LL n, LL c)
{
    LL i = , k = ;
    LL x = rand() % (n - ) + ;
    LL y = x;
    while(true)
    {
        i++;
        x = (multi(x, x, n) + c) % n;
        LL d = gcd((y - x + n) % n, n);
        if( < d && d < n) return d;
        if(y == x) return n;
        if(i == k)
        {
            y = x;
            k <<= ;
        }
    }
}

void find(LL n, int c)
{
    if(n == ) return;
    if(Miller_Rabin(n))
    {
        fac[ct++] = n;
        return ;
    }
    LL p = n;
    LL k = c;
    while(p >= n) p = pollard_rho(p, c--);
    find(p, k);
    find(n / p, k);
}

int main()
{
    LL n,kind;
    scanf("%lld%lld",&kind,&n);
    ct = ;
    find(n, );
    sort(fac, fac + ct);
    num[] = ;
    int k = ;
    for(int i=; i<ct; i++)
    {
        if(fac[i] == fac[i-])
            ++num[k-];
        else
        {
            num[k] = ;
            fac[k++] = fac[i];
        }
    }
    cnt = k;
    LL ans=(1ll<<);
    for(int i=;i<cnt;i++){
        LL as=,N=kind/fac[i];
        while(N){
            as+=N;
            N/=fac[i];
        }
        as/=num[i];
        ans=min(as,ans);
    }
    if(ans==1ll<<)ans=;
    printf("%lld",ans);
    return ;
}