Problem 500!!!

The number of divisors of 120 is 16.
In fact 120 is the smallest number having 16 divisors.

Find the smallest number with 2500500 divisors.
Give your answer modulo 500500507.

前500500個質數是第一個假的解，然後維護一個"亂七八糟的資料結構"裡面存著第一個質數的平方，第二個質數的平方，...，然後每次把這個資料結構裡最小的那個和當前解最大的單冪次質數做個比較，如果那個元素更小就替換，然後在資料結構裡推進剛剛用來替換的那個形如p^k的數的平方......大概就這麼個思路吧...

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 500500507;
const ll mi = 500500+1;
const ll maxn = 2e7;//從2開始，到第500500個素數是7e6左右
ll a[4] = {2, 4, 8, 16};
ll prime[mi+100];
ll check[maxn], tot;
void init(){
    memset(check, 0, sizeof(check));
    prime[tot++] = 1;
    for (ll i = 2; i < maxn; ++i){
        if(!check[i]){
            prime[tot++] = i;
        }
        if(tot == mi){
            break;
        }
        for (ll j = 1; j < tot; ++j){
            if (i * prime[j] > maxn){
                break;
            }
            check[i*prime[j]] = 1;
            if (i % prime[j] == 0){
                break;
            }
        }
    }
}
//快速冪，返回a^b
ll Pow(ll a,ll b)
{
    ll ret=1;
    while(b){
        if(b&1) ret=(ret*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return ret;
}

int main(){
    init();
    priority_queue<ll>que;
    que.push(1);
    ll Size = 1;
    ll Top = 4;
    for(ll i = 1; i < mi; i++){
        if(Size >= mi){
            if(prime[i] >= que.top()){
                break;
            }
            else{
                que.pop();
                que.push(prime[i]);
                for(ll j = 0; j < Top; j++){
                    if(Pow(prime[i], a[j]) >= prime[tot-1]){
                        Top = j;
                        break;
                    }
                    if(Pow(prime[i], a[j]) >= que.top()){
                        Top = j;
                        break;
                    }
                    else{
                        que.pop();
                        que.push(Pow(prime[i], a[j]));
                    }
                }
            }
        }
        else{
            que.push(prime[i]);
            Size++;
            for(ll j = 0; j < Top; j++){
                if(Pow(prime[i], a[j]) >= prime[tot-1]){
                    Top = j;
                    break;
                }
                if(Size >= mi){
                    if(Pow(prime[i], a[j]) >= que.top()){
                        Top = j;
                        break;
                    }
                    else{
                        que.pop();
                        que.push(Pow(prime[i], a[j]));
                    }
                }
                else{
                    if(Pow(prime[i], a[j]) >= prime[tot-1]){
                        Top = j;
                        break;
                    }
                    que.push(Pow(prime[i], a[j]));
                    Size++;
                }
            }
        }
    }
    ll ans = 1;
    while(!que.empty()){
        ans *= que.top();
        que.pop();
        ans%=mod;
    }
    cout<<ans<<endl;
    return 0;
}
 

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 500500507;
const ll N = 500500;
const ll L =N*16;
ll np[L+1];

struct xpowy{
    ll x,y;
    bool operator <(xpowy &b){
        return log(x)*(y+1)>log(b.x)*(b.y+1);
    }
};
//快速冪，返回a^b
ll Pow(ll a,ll b)
{
    ll ret=1;
    while(b){
        if(b&1) ret=(ret*a)%mod;
        a=(a*a)%mod;
        b>>=1;
    }
    return ret;
}

int main(){
    vector<xpowy> v;
    for(ll n=2;n<=L;n++)
    if(!np[n]){
        for(ll x=n*n;x<=L;x+=n)
            np[x]=1;
        v.push_back({n,0});
    }
    make_heap(v.begin(),v.end());
    for(int i=0;i<N;i++){
        v[0].y=v[0].y*2+1;
        pop_heap(v.begin(),v.end());
        push_heap(v.begin(),v.end());
    }
    ll s=1;
    for(int i=0;i<v.size();i++)
        s=s*Pow(v[i].x,v[i].y)%mod;
    cout<<s;
}