HDU 5428 The Factor（分解质因子）_5428的因数-优快云博客

本文介绍了一个算法问题，即求解一系列正整数乘积的最小合数因子，该因子至少包含三个质因子。通过分解每个输入数的质因子并找出最小的两个质因子来解决此问题。提供了一段C++实现代码，包括对每个数进行质因子分解的过程。

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

2 3 1 2 3 5 6 6 6 6 6

Sample Output

6 4

<span style="font-family: "Courier New", Courier, monospace;">题意：求一列数的乘积的最小的两个质因子的积，如果积没有两个及以上个质因子，输出-1.</span>

思路：对每个数分解质因子，保存最小的两个质因子,可以用短除法，从2——sqrt（这个数）.

本鶸：CE：1次 TLE：2次 RE：1次

TLE是因为没有从2——sqrt（这个数），而是直接2——这个数......开始还以为是短除法用错了，就打了个素数表，每次除一个素数，试了一下还是TLE，主要还是前面那个原因.

AC代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long ll;
int a[310],sizee;
int main()
{
    int T,n,m;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        memset(a,0,sizeof a);
        sizee=0;
        while(n--)
        {
            int flag=0;
            scanf("%d",&m);
            for(int i=2;i<=sqrt(m+0.5);i++)
            {
                while(m%i==0)
                {
                    m/=i;
                    a[sizee++]=i;
                    flag++;
                    if(flag==2)
                        break;
                }
                if(flag==2)
                    break;
            }
            if(flag!=2&&m!=1)
                    a[sizee++]=m;
        }
        if(sizee<2)
            cout<<-1<<endl;
        else
        {
            sort(a,a+sizee);
            cout<<(ll)a[0]*a[1]<<endl;
        }
    }
    return 0;
}