hdu 5428 The Factor 找质因子

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本文介绍了一个算法问题，即在一个整数序列中找到所有元素乘积的最小复合因子（该因子本身具有超过两个的因子）。文章提供了完整的源代码实现，并解释了核心思路：通过分解每个整数来找出其最小的两个质因子，然后返回这两个质因子的乘积。

The Factor

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1827 Accepted Submission(s): 551

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

T answers in

T lines.

Sample Input

Sample Output

6
4

Source

BestCoder Round #54 (div.2)

求最小的两个质因子乘积即可

<span style="font-family: Arial, Helvetica, sans-serif;">#include<iostream></span>#include<cstdio>
#include<map>
using namespace std;
#define ll long long
ll inf = 100000000007ll;
int main(){
    int t,n;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        ll u;
        ll x1=inf,x2=inf;
        for(int j = 0;j < n; j++){
            scanf("%lld",&u);
            for(ll i = 2;i * i <= u; i++){
                while(u % i == 0){
                    if(i < x1) x2 = x1, x1 = i;
                    else if(i < x2) x2 = i;
                    u /= i;
                }
            }
            if(u != 1){
                if(u < x1) x2 = x1, x1 = u;
                else if(u < x2) x2 = u;
            }
        }
        if(x2 == inf) printf("-1\n");
        else printf("%lld\n",x1*x2);
    }
    return 0;
}