【笨方法学PAT】1085 Perfect Sequence （25 分）

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本文介绍了一个算法问题，即如何从给定的正整数序列中找到尽可能多的数，形成一个完美子序列，其中序列的最大值不超过最小值乘以参数p。文章提供了问题的输入输出规格，样例输入输出，并分享了高效解决该问题的代码实现。

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一、题目

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 mare 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:

二、题目大意

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。。

三、考点

逻辑

四、注意

1、直接遍历会超时；

2、使用int会有一个点报错，要使用long long；

3、维护一个最长length，剪枝减小时间复杂度。

五、代码

#include<iostream>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
int main() {
	//read
	long long int n, p;
	cin >> n >> p;
	vector<int> v(n);
	for (int i = 0; i < n; ++i) {
		scanf("%d", &v[i]);
	}

	//sort
	sort(v.begin(), v.end());

	//solve

	//out of time
	/*
	int i;
	bool flag = false;
	for (i = 0; i < n; ++i) {
		long long int a, b;
		for (int j = 0; j <= i; ++j) {
			a = v[j];
			b = v[n - i + j - 1];
			if (b <= a * p) {
				flag = true;
				break;
			}
		}
		if (flag == true)
			break;
	}
	*/

	int result = 0;
	for (int i = 0; i < n; ++i) {
		for (int j = i + result; j < n; ++j) {
			if (v[j] <= v[i] * p)
				result = max(j - i + 1, result);
			else
				break;
		}
	}

	//output
	cout << result << endl;

	system("pause");
	return 0;
}