牛客小白月赛6 水题 求n!在m进制下末尾0的个数 数论

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
#define ls (r<<1)
#define rs (r<<1|1)
#define debug(a) cout << #a << " " << a << endl
using namespace std;
typedef long long ll;
const ll maxn = 1e6+10;
const double eps = 1e-8;
const ll mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
ll f[100]={-1,1,0,0,2,10,4,40,92,352,724,2680,14200,73712,365596};
ll 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 getcnt( ll p, ll x ) {
    ll res = 0;
    while(x) {
        res += x/p;
        x /= p;
    }
    return res;
}
int main() {
    ios::sync_with_stdio(0);
    ll a[105];
    a[1] = 1, a[2] = 1;
    for( ll i = 3; i <= 92; i ++ ) {
        a[i] = a[i-1] + a[i-2];
    }
    ll x, m;
    cin >> x >> m;
    bool flag = false;
    for( ll i = 1; i <= 92; i ++ ) {
        if( a[i] == x ) {
            flag = true;
            break;
        }
    }
    if( flag ) {
        map<ll,ll> mp;
        vector<pair<ll,ll> > e;
        for( ll i = 1; i <= 25; i ++ ) {
            while(m%prime[i]==0) {　　//m中有多个相同的质数
                mp[prime[i]] ++;
                m /= prime[i];
            }
        }
        for( auto i : mp ) {
            e.push_back(make_pair(i.second,getcnt(i.first,x)));
        }
        ll k = 1e18+1;
        for( ll i = 0; i < e.size(); i ++ ) {
            k = min(k,e[i].second/e[i].first);　　//因为质数可能有多个，所以求的质数还要除以质数的个数
        }
        cout << k << endl;
    } else {
        cout << f[x%min((ll)13,m+1)+1] << endl;
    }
    return 0;
}