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

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

Print T answers
in T lines.



Sample Input

2
3
1 2 3
5
6 6 6 6 6

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>