BestCoder Round #54 (div.2) HDOJ5428 The Factor(脑洞)

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本文介绍了一道算法题目，需要找出一系列正整数乘积中大于2个因子的最小因子。通过分解质因数的方法解决该问题，并给出AC代码实现。

The Factor

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 495 Accepted Submission(s): 154

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

每个数有用的是自己的质因子，求出质因子并保存，最小的两个乘积即为答案。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int MAXN = 110;
typedef long long ll;
int n;
ll a[MAXN], ans[MAXN << 2];
int main(int argc, char const *argv[])
{
	int t;
	scanf("%d", &t);
	while(t--) {
		int cnt = 0;
		scanf("%d", &n);
		for(int i = 1; i <= n; ++i) {
			scanf("%lld", &a[i]);
			for(int j = 2; (ll)j * j <= a[i]; ++j)
				while(a[i] % j == 0) {
					ans[cnt++] = j;
					a[i] /= j;
				}
			if(a[i] != 1) ans[cnt++] = a[i];
		}
		if(cnt < 2) printf("-1\n");
		else {
			sort(ans, ans + cnt);
			printf("%lld\n", ans[0] * ans[1]);
		}
	}
	return 0;
}</span>