Codeforces 762A-k-th divisor

最新推荐文章于 2024-11-18 21:13:50 发布

原创最新推荐文章于 2024-11-18 21:13:50 发布 · 269 阅读

0 ·

CC 4.0 BY-SA版权

----其他同时被 2 个专栏收录

216 篇文章

订阅专栏

Codeforces

140 篇文章

订阅专栏

本文介绍了一种求解特定整数第K大因子的高效算法，并提供了完整的C++实现代码。该算法适用于数据范围较大的场景，通过双vector存储因子的方式优化了查找效率。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

题意：求一个数第k大的因子，不存在第k大的因子则输出-1

解题思路：因为数据较大，用两个vector来存因子

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
#include <set>
#include <stack>
#include <map>
#include <climits>

using namespace std;

#define LL long long

int main()
{
    LL n,k;
    while(~scanf("%lld %lld",&n,&k))
    {
        vector <LL> v1,v2;
        for(LL i=1; i*i<=n; i++)
        {
            if(n%i==0)
            {
                v1.push_back(i);
                if(i*i!=n) v2.push_back(n/i);
            }
        }
        LL len=v1.size()+v2.size(),len1=v1.size(),len2=v2.size();
        if(k>len) printf("-1\n");
        else
        {
            if(k<=len1) printf("%lld\n",v1[k-1]);
            else printf("%lld\n",v2[len2-(k-len1)]);
        }
    }
    return 0;
}