Codeforces 762A-k-th divisor

k-th divisor time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

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; }