1085 Perfect Sequence (25)

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本文介绍了一种算法，用于从给定的正整数序列中寻找最长的完美子序列。完美子序列定义为该子序列的最大值不超过最小值乘以给定参数p。文章通过排序和滑动窗口的方法实现了这一目标。

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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 (<= 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:

思路：

给定一个正整数数列，和正整数p，设这个数列中的最大值是M，最小值是m，如果M <= m * p，则称这个数列是完美数列。现在给定参数p和一些正整数，请你从中选择尽可能多的数构成一个完美数列。输入第一行给出两个正整数N（输入正数的个数）和p（给定的参数），第二行给出N个正整数。在一行中输出最多可以选择多少个数可以用它们组成一个完美数列

C++：

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main() {
    int n;
    long long p;
    scanf("%d%lld", &n, &p);
    vector<int> v(n);
    for (int i = 0; i < n; i++)
        cin >> v[i];
    sort(v.begin(), v.end());
    int result = 0, temp = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i + result; j < n; j++) {
            if (v[j] <= v[i] * p) {
                temp = j - i + 1;
                if (temp > result)
                    result = temp;
            } else {
                break;
            }
        }
    }
    cout << result;
    return 0;
}