题目地址

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 (≤10
​5
​​ ) is the number of integers in the sequence, and p (≤10
​9
​​ ) is the parameter. In the second line there are N positive integers, each is no greater than 10
​9
​​ .

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<algorithm>
#include <cmath>
#include<map>
#include<cstring>
#include<queue>
#include<string>
#include<set>
#include<stack>
using namespace std;
typedef long long ll;
const int maxn=100010,inf=100000000;
int n,ans=0;ll p,a[maxn];
int search(ll x,int l,int r){           //本质要查第一个比x大的数ans；ans-1就是完美序列最后一个数；
    if(x>=a[r]) return n;
    while(l<r){
        int mid=(l+r)/2;
        if(a[mid]>x) r=mid;
        else l=mid+1;
    }
    return l;
}
int main(){
    cin>>n>>p;
    for(int i=0;i<n;i++) cin>>a[i];
    sort(a,a+n);
    int left=0;
    for(int i=0;i<n;i++){
        int j=search(a[i]*p,i,n-1);
        ans=max(ans,j-i);
    }
    cout<<ans<<endl;
}