A. 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.
1 //2017.02.01 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 6 using namespace std; 7 8 int divisor[35000000]; 9 10 int main() 11 { 12 long long n, k; 13 while(cin>>n>>k) 14 { 15 long long ans = -1; 16 int cnt = 0; 17 bool fg = false; 18 for(long long i = 1; i*i <= n; i++) 19 { 20 if(n%i==0)divisor[++cnt] = i; 21 if(i*i == n)fg = true; 22 } 23 if(fg && k > 2*cnt-1)cout<<-1<<endl; 24 else if(k > 2*cnt)cout<<-1<<endl; 25 else if(k<=cnt)cout<<divisor[k]<<endl; 26 else{ 27 if(!fg)cout<<(n/divisor[2*cnt+1-k])<<endl; 28 else cout<<(n/divisor[2*cnt-k])<<endl; 29 } 30 } 31 32 return 0; 33 }