1085 Perfect Sequence (25 分) 双指针， 短小精悍

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

Note

1. 题意：M≤m×p find最大长度，其中m位最小值，M为最大值
2. 思路：亮点 J 从 cnt 开始即可，因为目的是找到更长的cnt

Code

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
long long a[100010] = {0};
int main(){
    int num, p;
    int cnt = 0;
    cin >> num >>p;
    for(int i = 0; i < num; i++){
        cin >> a[i];
    }
    sort(a, a+num);
    for(int i = 0; i < num; i++){
        for(int j = i + cnt; j < num;j++ ){
            if(a[i] * p >= a[j] ){
                if(j - i + 1 > cnt) cnt = j - i + 1;
            }
            else break;
        }
    }
    cout << cnt;
	return 0;
}