输入n，m求n!在m进制下末尾0的个数。
n, m<1e18

图片说明 

#include<bits/stdc++.h>
#define LL long long
using namespace std;

int prime[]={0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97};
LL ind[105], cnt[105];

LL getcnt(LL x, LL n){//计算1-n中所有质因子x的总和个数
    LL res=0;
    while(n){
        res+=n/x;
        n/=x;
    }
    return res;
}

int main(){

    LL n, m;
    scanf("%lld%lld", &n, &m);
    for(int i=1; i<=25; i++){
        LL mm=m;
        while(mm%prime[i]==0){
            ind[prime[i]]++;

            mm/=prime[i];
        }
    }
    for(int i=1; i<=25; i++){
        if(ind[prime[i]]){
            cnt[prime[i]]=getcnt(prime[i], n);
        }
    }
    LL ans=1e18+1;
    for(int i=1; i<=25; i++){
        if(ind[prime[i]]){
            ans=min(ans, cnt[prime[i]]/ind[prime[i]]);
        }
    }
    cout<<ans<<endl;

    return 0;
}