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

#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
    long long len,p,i,tem,j;
    vector<long long> vv;
    scanf("%lld%lld",&len,&p);
    for(i = 0;i<len;i++)
    {
        scanf("%lld",&tem);
        vv.push_back(tem);
    }

    sort(vv.begin(),vv.end());

    int Max= 0;
    int count;
    i = 0; j = 0;
    for(i = 0 ; i < len ;i ++)
    {
        j = i + Max;
        count = Max;
        while( j < len && vv[j] <= vv[i] * p)
        {
            ++j;
            ++ count;
        }
        if(Max < count) Max = count;
    }
    printf("%d
",Max);
    return 0;
}