标签：text cto rem math 代码实现 iter title i++ 代码
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.
Example
4 2
2
5 3
-1
12 5
6
代码实现：
1 #include<cstdio> 2 #include <set> 3 #include<vector> 4 #include<iostream> 5 #include<algorithm> 6 #include<cmath> 7 using namespace std; 8 int main() 9 { 10 set<long long int>a; 11 long long int n,k; 12 long long int p1,pp; 13 scanf("%lld%lld",&n,&k); 14 pp=sqrt(n); 15 for(int i=1; i<=pp; i++) 16 { 17 p1=n%i; 18 if(p1==0) 19 { 20 a.insert(n/i); 21 a.insert(i); 22 } 23 } 24 vector<long long int>v; 25 insert_iterator<vector<long long int> > in_it(v, v.begin()); 26 copy(a.begin(), a.end(), in_it); 27 //printf("%d\n",v.size()); 28 if(k>v.size()) 29 printf("-1\n"); 30 31 else 32 { 33 printf("%lld\n", v[k-1]); 34 } 35 36 return 0; 37 }
标签：text cto rem math 代码实现 iter title i++ 代码
原文地址：http://www.cnblogs.com/2016024291-/p/7043896.html
