PAT A1085. Perfect Sequence (25)

Given a sequence of positive integers and another positive integer p. The sequence is said to be a "perfect sequence" if M <= m * p where M and m are the maximum and minimum numbers in the sequence, respectively.

Now given a sequence and a parameter p, you are supposed to find from the sequence as many numbers as possible to form a perfect subsequence.

Input Specification:

Each input file contains one test case. For each case, the first line contains two positive integers N and p, where N (<= 10⁵) is the number of integers in the sequence, and p (<= 10⁹) is the parameter. In the second line there are N positive integers, each is no greater than 10⁹.

Output Specification:

For each test case, PRint in one line the maximum number of integers that can be chosen to form a perfect subsequence.

10 82 3 20 4 5 1 6 7 8 9Sample Output:
8
#include <cstdio>#include <algorithm>#include <cstring>#define Max 100010using namespace std;int main(){	int n,m,S[Max];	scanf("%d%d",&n,&m);	for(int i=0;i<n;i++)	{		scanf("%d",&S[i]);	}	sort(S,S+n);	int mm=1,f;	for(int i=0;i<n;i++)	{		f=upper_bound(S+i+1,S+n,(long long )S[i]*m)-S;		f-=i;		mm=max(mm,f);	}	printf("%d/n",mm);	system("pause");	return 0;}