1085. 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.

Sample Input:

10 8
2 3 20 4 5 1 6 7 8 9

Sample Output:

8

#include<iostream>
#include<vector>
#include<cstdio>
#include<algorithm>
using namespace std;
int main(){
	int n,p;
	scanf("%d%d",&n,&p);
	vector<long>vt;
	int i,j,val;
	for(i=0;i<n;i++){
		scanf("%ld",&val);
		vt.push_back(val);
	}
	sort(vt.begin(),vt.end());
	int left=1,right=0;
	int length=0;
	int len=0;
	for(i=0;i<n;i++){
		long tmp = p*vt[i];
		if(tmp>=vt[n-1]){
			if(len < n-i){
				len=n-i;
			}
			break;
		}
		int right=n-1;
		while(right>left){
			int mid=(right+left)/2;
			if(vt[mid]>tmp){
				right=mid;
			}else if(vt[mid]<tmp){
				left=mid+1;
			}else {
				left=mid+1;
				break;
			}
		}
		if(left-i>len){
			len=left-i;
		}
	}
	printf("%d\n",len);
	return 0;
}