HihoCoder - 1789

https://cn.vjudge.net/problem/HihoCoder-1789 

题意：给定 n, k，求一个最大的整数 m，使得 km 是 n! 的约数

思路：将k素因子分解，看n!中存在k的素因子次方数最少的个数，改个数就是ans；最少的能除尽，那么比它多的也必然能除尽；

#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<map>
#include<stack>
#include<string>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-6
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
#define MOD(x) ((x%mod)+mod)%mod
#define FASTIO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
//ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=2e5+9;
const int mod=1e8+7;

template <class T>
inline void sc(T &ret)
{
    char c;
    ret = 0;
    while ((c = getchar()) < '0' || c > '9');
    while (c >= '0' && c <= '9')
    {
        ret = ret * 10 + (c - '0'), c = getchar();
    }
}

ll p_fac[109],expo[109],prime[maxn/10],prime_num;
bool is_prime[maxn];
void init_prime()
{
    mem(is_prime,1);
    for(int i=2;i<maxn;i++)
    {
        if(is_prime[i])
        {
            prime[prime_num++]=i;
            for(int j=i+i;j<maxn;j+=i) is_prime[j]=0;
        }
    }
}
int main()
{
    //FASTIO;
    //#define endl '\n'
    ll n,k;
    cin>>n>>k;
    init_prime();
    ll ans=1e18;
    int prime_fac=0;
    for(int i=0;i<prime_num&&prime[i]*prime[i]<=k;i++)
    {
        if(k%prime[i]==0)
        {
            expo[prime_fac]=0;
            while(k%prime[i]==0)
            {
                expo[prime_fac]++;
                k/=prime[i];
            }
            p_fac[prime_fac++]=prime[i];
        }
    }
    if(k!=1)
    {
        expo[prime_fac]=1;
        p_fac[prime_fac++]=k;
    }
    //分解n!
    for(int i=0;i<prime_fac;i++)
    {
        ll n_exp=0;
        ll t=n;
        while(t)
        {
            n_exp+=t/p_fac[i];
            t/=p_fac[i];
        }
        ans=min(ans,n_exp/expo[i]);
    }
    cout<<ans<<endl;
}