1085 Perfect Sequence （25 分）

最新推荐文章于 2024-08-25 16:53:36 发布

SourDumplings

最新推荐文章于 2024-08-25 16:53:36 发布

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本文介绍了一种使用简单动态规划法解决寻找最长完美子序列的问题。对于给定的一系列正整数和参数p，目标是找到尽可能多的数形成一个完美子序列，即序列中最大值不超过最小值乘以p。通过排序和滑动窗口技术，文章提供了一个有效的算法实现。

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 (≤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:

C++:

简单动态规划法：

/*
 @Date    : 2018-02-16 17:10:13
 @Author  : 酸饺子 (changzheng300@foxmail.com)
 @Link    : https://github.com/SourDumplings
 @Version : $Id$
*/

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

#include <iostream>
#include <cstdio>
#include <algorithm>

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;
}