A1085

1085 Perfect Sequence (25分)

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<string>
#include<vector>
#include<map>
#include<unordered_map> 
#include<set>
#include<cstdio>
#include<algorithm>
using namespace std;
int main() {
    int n;
    long long p;
    scanf("%d%lld", &n, &p);
    vector<int> v(n);
    for (int i = 0; i < n; i++)
        cin >> v[i];
    sort(v.begin(), v.end());
    int result = 0, temp = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + result; j < n; j++) {
            if (v[j] <= v[i] * p) {
                temp = j - i + 1;
                if (temp > result)
                    result = temp;
            } else {
                break;
            }
        }
    }
    cout << result;
    return 0;
}