PAT甲级——A1085 Perfect Sequence

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 (≤) is the number of integers in the sequence, and p (≤) is the parameter. In the second line there are N positive integers, each is no greater than 1.

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

又是没看清题，这道题的子序列不需要是原来的连续子序列，只要求是原来里面的值就行，搞得又浪费了很多时间！！！！

//靠，不需要是子排序，就是找数字就行
#include <iostream>
#include <deque>
#include <vector>
#include <algorithm>
using namespace std;
int N;
long long P;
int main()
{
    cin >> N >> P;
    vector<int>num(N);
    for (int i = 0; i < N; ++i)
        cin >> num[i];
    sort(num.begin(), num.end());
    int res = 0;
    for(int L=0,R=0;L<=R && R<N;++R)
    {
        while (L <= R && num[R] > P * num[L])
            L++;
        res = res > R - L + 1 ? res : R - L + 1;        
    }
    cout << res << endl;
    return 0;
}