Almost All Divisors（求因子个数及思维）

梦里梦外; 2021-10-29 15:18 180阅读 0赞

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We guessed some integer number xx. You are given a list of almost all its divisors. Almost all means that there are all divisors except 11and xx in the list.

Your task is to find the minimum possible integer xx that can be the guessed number, or say that the input data is contradictory and it is impossible to find such number.

You have to answer tt independent queries.

Input

The first line of the input contains one integer tt (1≤t≤251≤t≤25) — the number of queries. Then tt queries follow.

The first line of the query contains one integer nn (1≤n≤3001≤n≤300) — the number of divisors in the list.

The second line of the query contains nn integers d1,d2,…,dnd1,d2,…,dn (2≤di≤1062≤di≤106), where didi is the ii-th divisor of the guessed number. It is guaranteed that all values didi are distinct.

Output

For each query print the answer to it.

If the input data in the query is contradictory and it is impossible to find such number xx that the given list of divisors is the list of almost allits divisors, print \-1. Otherwise print the minimum possible xx.

Example

input

Copy

2
    8
    8 2 12 6 4 24 16 3
    1
    2

output

Copy

48
    4

思路：求出因子个数，看是否这n个数是否包含这n个因子数，然后判断一下再判断一下这n个数是否是他的因子

代码：

#include<cstdio>
    #include<iostream>
    #include<cstring>
    #include<algorithm>
    #include<queue>
    #include<stack>
    #include<set>
    #include<vector>
    #include<map>
    #include<cmath>
    const int maxn=1e5+5;
    typedef long long ll;
    using namespace std;
    ll count(ll n){
        ll s=1;
        for(ll i=2;i*i<=n;i++){
            if(n%i==0){
                int a=0;
                while(n%i==0){
                    n/=i;
                    a++;
                }
                s=s*(a+1);
            }
        }
        if(n>1) s=s*2;
        return s;
    }
    ll a[maxn];
    int main()
    {
       int T;
       cin>>T;
       int n;
       while(T--)
       {
           scanf("%d",&n);
           ll maxx=2;
           ll minn=10000000;
           ll x;
           for(int t=0;t<n;t++)
           {
               scanf("%lld",&a[t]);
               maxx=max(a[t],maxx);
               minn=min(a[t],minn);
        }
        ll ans=maxx*minn;
        bool flag=false;
        for(int t=0;t<n;t++)
        {
            if(ans%a[t]!=0)
            {
                flag=true;
            }
        }
        if(count(ans)-2==n&&flag==false)
        printf("%lld\n",ans);
        else
        {
            printf("-1\n");
        }
        
    
       }
       return 0;
    }

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转载于:https://www.cnblogs.com/Staceyacm/p/10869742.html