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×pwhere 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

C++:

简单动态规划法：

/*
 @Date    : 2018-02-16 17:10:13
 @Author  : 酸饺子 ([email protected])
 @Link    : https://github.com/SourDumplings
 @Version : $Id$
*/

/*
https://www.patest.cn/contests/pat-a-practise/1085
 */

#include 
#include 
#include 

using namespace std;

int main(int argc, char const *argv[])
{
    int N, p;
    scanf("%d %d", &N, &p);
    long long num[N];
    for (int i = 0; i != N; ++i)
        scanf("%lld", &num[i]);
    sort(num, num+N);
    int maxl = 0;
    int l, last_s = N - 1;
    for (int e = N - 1; e != -1; --e)
    {
        int s;
        if (last_s > 0 && num[e] > num[last_s-1] * p)
            continue;
        for (s = last_s - 1; s != -1; --s)
        {
            if (num[e] > num[s] * p)
                break;
        }
        l = e - s;
        last_s = s + 1;
        if (l > maxl)
            maxl = l;
    }
    printf("%d\n", maxl);
    return 0;
}