B1030 perfect sequence

1030 Perfect sequence (25 branch )

Given a sequence of positive integers , And positive integers p, Let the maximum of this sequence be M, The minimum is m, If M≤mp, We call this sequence perfect .

Now, given the parameters p And some positive integers , Please choose as many numbers as you can to form a perfect sequence .

Input format ：

The first line of input gives two positive integers N and p, among N（≤10 Of 5 Power ） Is the number of positive integers entered ,p（≤10 Of 9 Power ） Is the given parameter . The second line gives N A positive integer , No more than 10 Of 9 Power .

Output format ：

How many numbers can you choose to output in a row? You can use them to form a perfect sequence .

sample input

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

sample output

8

Topic analysis ：

1. After storing the array data , Use it in time sort Sort , Convenient for future operation
2. Test point 4 Timeout problem ： You can save a temporary variable first j,j Is the maximum number in the current sequence . finish while Update in time after the cycle count Value
3. Be careful mp The numerical range of

The code is as follows ：

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n,p,count=0;
    cin>>n>>p;
    vector<int> a;
    for(int i=0;i<n;i++){
        int x;
        cin>>x;
        a.push_back(x);
    }
    sort(a.begin(),a.end());
    int j=0;// Time spent optimizing code ------ Test point 4 The key to timeout 
    for(int i=0;i<n-count;i++){
        long long int mp=(long long)a[i]*p;// Take care mp Value int There is no room for ------- Test point 5
        
        while(j<n&&a[j]<=mp){
            j++;
        }
        count=max(count,j-i);
    }
    cout<<count;
    
    return 0;
}