PAT.A1085. Perfect Sequence

最新推荐文章于 2020-06-08 15:04:44 发布

克拉拉123

最新推荐文章于 2020-06-08 15:04:44 发布

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本文链接：https://blog.youkuaiyun.com/mokena1/article/details/79955294

本文介绍了一个算法问题：如何从给定的整数序列中找出尽可能多的数字形成一个完美子序列。其中，完美子序列定义为序列中的最大值不超过最小值乘以给定参数p。文章提供了完整的C++实现代码，并通过样例展示了算法的有效性。

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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 (<= 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:

#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <iostream>
using namespace std;


long long a[100010];
int main(){
	int n, p;
	scanf("%d%d", &n, &p);
	for (int i = 0; i < n; i++)
		scanf("%lld", &a[i]);
	sort(a, a + n);
	int count = -1;
	int i = 0, j = 0;
	while (j <n) {
		
		while (a[i] * p>=a[j]&&j<n) j++;
	//	printf("%lld %d %d\n", a[j],i,j);
		count = max(count, j - i);
		i++;
	}
	printf("%d",count);
	return 0;
	
}