2019ICPC徐州區域賽 E.Multiply

鏈接

點擊跳轉

題解

$\sum a_i < Y$也就意味着$\prod a_i! < Y$

所以答案至少是$0$

那麼我只需要對$X$分解質因數，然後看一下每個質因數還能允許我容納幾個$x$就可以了

分解質因數用Pollard Rho算法

代碼

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct Pollard_Rho
{
    ll md(ll x, ll p){while(x>p)x-=p;return x;}
    ll mult(ll a, ll b, ll p)
    {
        ll ans=0, t=a;
        for(;b;b>>=1,t=md(t+t,p))if(b&1)ans=md(ans+t,p);
        return ans;
    }
    ll pow(ll a, ll b, ll p)
    {
        ll ans=1, t=a;
        for(;b;b>>=1,t=mult(t,t,p))if(b&1)ans=mult(ans,t,p);
        return ans;
    }
    bool MR(ll n)
    {
        ll a, t, u, i, x, y, tim=20;
        if(n==2)return true;
        if(~n&1)return false;
        if(n==3 or n==5)return true;
        if(n%3==0 or n%5==0)return false;
        for(t=0,u=n-1;~u&1;u>>=1,t++);
        while(tim--)
        {
            a=rand()%(n-1)+1;
            for(y=x=pow(a,u,n),i=1;i<=t;i++)
            {
                x=mult(x,x,n);
                if(x==1 and y!=n-1 and y!=1)return false;
                y=x;
            }
            if(x!=1)return false;
        }
        return true;
    }
    ll pollard_rho(ll n, ll c)
    {
        ll i, k, x=rand()%n, y=x, d;
        for(i=1,k=2;;i++)
        {
            x=(mult(x,x,n)+c)%n;
            d=__gcd(y>x?y-x:x-y,n);
            if(d>1 and d<n)return d;
            if(x==y)return n;
            if(i==k)y=x,k<<=1;
        }
    }
    void find(ll n, ll c, map<ll,ll>& ans)
    {
        if(n==1)return;
        if(MR(n)){ans[n]++;return;}
        ll p=pollard_rho(n,c);
        for(;p==n;c++)p=pollard_rho(n,c);
        find(p,c,ans), find(n/p,c,ans);
    }
    map<ll,ll> run(ll n)
    {
        map<ll,ll> ans;
        find(n,1,ans);
        return ans;
    }
};
int main()
{
    ll T=read();
    Pollard_Rho PR;
    while(T--)
    {
        ll N=read(), X=read(), Y=read(), i;
        vector<ll> a(N+5);
        rep(i,1,N)a[i]=read();
        auto lis = PR.run(X);
        ll ans=linf;
        for(auto pr:lis)
        {
            ll cnt=0;
            rep(i,1,N)
            {
                ll t=a[i], p=0;
                while(t)
                {
                    p += t/pr.first;
                    t/=pr.first;
                }
                cnt -= p;
            }
            ll t=Y, p=0;
            while(t)
            {
                p += t/pr.first;
                t/=pr.first;
            }
            cnt += p;
            ans = min(ans,cnt/pr.second);
        }
        printf("%lld\n",ans);
    }
    return 0;
}