PAT 1085. Perfect Sequence (25)

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

 将序列的左端固定，然后用二分法确定最长的序列的右端。然后对左端的进行循环，确定最长的序列。

#include <iostream>
#include <algorithm>

using namespace std;

long long sequence[100000];
long long p;

int main()
{
    long long n;
    cin >> n >> p;
    for (int i = 0; i < n; i++)
        cin >> sequence[i];
    sort(sequence, sequence + n);
    int maxLength = 0;
    for (int i = 0; i < n; i++)
    {
        if (maxLength >= n - i)
            break;
        int left = i, right = n - 1;
        if (sequence[i] * p >= sequence[right] && maxLength < n - i)
            maxLength = n - i;
        else
        {
            while (right - left != 1)
            {
                int middle = (left + right) / 2;
                if (sequence[i] * p >= sequence[middle])
                    left = middle;
                else
                    right = middle;
            }
            if (right - i > maxLength)
                maxLength = right - i;
        }
    }
    cout << maxLength;
39 }