Day 4

T1

上来思路就是暴力找出来每个数的只因数及其个数，然后暴力累乘

但是不对

只需要记每个数最多含多少个即可

做完就行

赛时写完了觉得自己很稳

结果挂分了，挂成了10

因为是要乘好多次同一个只因数，而不是只因数乘上那个次数

#include <bits/stdc++.h>
#define ll long long
#define gt getchar
#define pt putchar
using namespace std;
inline ll read(){
    ll x=0,f=1;char ch=gt();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=gt();}
    while(isdigit(ch)){x=(x<<1)+(x<<3)+(ch^48);ch=gt();}
    return x*f;}  
inline void print(ll x){if(x<0)pt('-'),x=-x;if(x>9)print(x/10);pt(x%10+48);}
void file(){freopen("buy.in", "r", stdin);freopen("buy.out", "w", stdout);}
const int Mod = 1e9 + 7;
int n;
int cnt[100005], bj[100005], tot;
map<int, bool> mp;
void r(int x) {
    for (int i = 2; i * i <= x; i++)
    {
        if (x % i == 0)
        {
            int tim = 0;
            while (x % i == 0) x /= i, ++ tim;
            if(i <= 100000)cnt[i] = max(cnt[i], tim);
            else bj[++ tot] = i;
        }
    }
    if (x != 1)
    {
        if(x <= 100000)cnt[x] = max(cnt[x], 1);
        else if(!mp[x])bj[++ tot] = x, mp[x] = 1;
    }
}
int main()
{
    //file();
    n = read();
    for(int i = 1; i <= n; ++ i)r(read());
    ll res = 1;
    for(int i = 1; i <= 100000; ++ i)
    {
        for(int j = 1; j <= cnt[i]; ++ j)
        {
            res = (res * i) % Mod;
        }
    }
    for(int i = 1; i <= tot; ++ i)
    {
        res = (res * bj[i]) % Mod;
    }
    cout << res << '\n';
    return 0;
}