HDU5428------The Factor

The Factor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1005    Accepted Submission(s): 325

提示：找出所有数的质因子，排序取其最小的两个；

Problem Description

There is a sequence of n positive integers. Fancycoder is addicted to learn their product, but this product may be extremely huge! However, it is lucky that FancyCoder only needs to find out one factor of this huge product: the smallest factor that contains more than 2 factors（including itself; i.e. 4 has 3 factors so that it is a qualified factor）. You need to find it out and print it. As we know, there may be none of such factors; in this occasion, please print -1 instead.

 

Input

The first line contains one integer T (1≤T≤15), which represents the number of testcases.

For each testcase, there are two lines:

1. The first line contains one integer denoting the value of n (1≤n≤100).

2. The second line contains n integers a1,…,an (1≤a1,…,an≤2×109), which denote these n positive integers.

 

Output

Print T answers in T lines.

 

Sample Input

231 2 356 6 6 6 6

 

Sample Output

64

 

Source

BestCoder Round #54 (div.2)

 

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#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<stack>#include<map>#include<set>#include<vector>#include<climits>#include<iomanip>#define LL long long#define uing unsigned int#define uLL unsigned LLusing namespace std;const int INF =0x3f3f3f3f;const double PI = acos(-1.0);const double esp=1e-6;const int N=5e4+10;LL prime[N>>2];bool flag[N];void getprime(){    int k=0;    for(int i=2; i<N; i++)    {        if(!flag[i])        {            prime[++k]=i;        }        for(int j=1; j<=k&&prime[j]*i<N; j++)        {            flag[i*prime[j]]=1;            if(i%prime[j]==0)                break;        }    }}LL a[N+10];vector<LL>B;bool cmp(LL x,LL y){    return x<y;}int main(){    getprime();    int t,n;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        {            B.clear();            for(int i=0; i<n; i++)            {                cin>>a[i];            }            for(int i=0; i<n; i++)            {                for(int j=1; prime[j]*prime[j]<=a[i]; j++)                {                    while(a[i]%prime[j]==0)                    {                        a[i]/=prime[j];                        B.push_back(prime[j]);                    }                }                if(a[i]!=1)                    B.push_back(a[i]);            }            if(B.size()<=1)                printf("-1\n");            else            {                sort(B.begin(),B.end(),cmp);                cout<<B[0]*B[1]<<endl;            }        }    }    return 0;}