codeforces762A

codeforces762A

You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn’t exist.

Divisor of n is any such natural number, that n can be divided by it without remainder.

Input
The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).

Output
If n has less than k divisors, output -1.

Otherwise, output the k-th smallest divisor of n.

Examples
Input
4 2
Output
2
Input
5 3
Output
-1
Input
12 5
Output
6
Note
In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.

In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn’t exist, so the answer is -1.

此处使用了vector(向量)，作为前面的因子以及后面的因子。
题目大意：找到给定n的第k个最小的因子。
求得一个因子后，可以求得另一个因子。

#include <iostream>
#include<stdio.h>
#include<vector>
using namespace std;
typedef long long LL;

int main()

{
	LL  n,k; int ans[1005];
	LL i, j;
	cin >> n >> k;
	vector <LL> v1, v2;
	for (LL i = 1; i*i <= n; i++)//对称
	{
		if (n%i == 0)
		{
			v1.push_back(i);
			if (i*i != n)//求得一个因子后，可以求得另一个因子。
				v2.push_back(n / i);//另一个对应的数
		}
	}

	if (v1.size()+v2.size() < k)
		cout << -1 << endl;
	else
	{
		if (k <= v1.size())
			cout << v1[k - 1] << endl;
		else
			cout << v2[v2.size() + v1.size() - k] << endl;//注意此处的-k关键，才能够保证从小到大
	}
	return 0;
}