【枚举约数】HackerRank - Week of Code 26 - Satisfactory Pairs

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
typedef vector<int>::iterator ITER;
vector<int>divisors[300010];
int n,used[300010],ans;
bool cmp(const int &a,const int &b)
{
	return a>b;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;++i)
	  {
	  	for(int j=1;j*j<=i;++j)
	  	  if(i%j==0)
	  		{
	  	      divisors[i].push_back(j);
	  	      if(j*j!=i)
	  	      	divisors[i].push_back(i/j);
	  	  	}
	  	sort(divisors[i].begin(),divisors[i].end(),cmp);
	  }
	for(int a=1;a<n;++a)//枚举a
	  for(int x=1;x*a<n;++x)//枚举x
		{
		  int yb=n-x*a;
		  for(ITER it=divisors[yb].begin();it!=divisors[yb].end();++it)//*it 就是b
		    {
		      if((*it)<=a)//因为b>a
		        break;
		      if(used[*it]!=a)//防止重复统计
		        {
		          ++ans;
		          used[*it]=a;
		        }
		    }
		}
	printf("%d\n",ans);
	return 0;
}