PAT甲级1085 Perfect Sequence (25 分)【双指针】_1085 perfect sequence (25分)双指针-优快云博客

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本文探讨了在给定序列和参数p的情况下，如何通过双指针算法找出满足M≤m×p条件的最大子序列长度，其中M和m分别是子序列的最大值和最小值。文章提供了详细的解题思路和参考代码。

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×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:

题目大意：

给出长度为n的序列以及一个参数p，要求选出一定的数字使M<= m * p，M为序列最大值，m为序列最小值，求最多能有多少个数字。

思路：

很明显是一道可以用双指针来解决的题目。

让两个指针都指向起始点，那么先固定I ，让j 动起来，找到最大的位置。随后再让i++，走向各大的最小数，此时j肯定又能够再变大即动起来。

参考代码：

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
int n,p;
vector<int> a;
int main(){
	scanf("%d%d",&n,&p);
	a.resize(n);
	for(int i = 0;i < n; ++i)	scanf("%d", &a[i]);
	sort(a.begin(), a.end());
	int i = 0, j = 0,cnt = 0; 
	while(i < n && j < n){
		while(a[i] * p >= a[j] && j < n)	j++;
		cnt = max(cnt , j - i);
		i++;
	}
	printf("%d", cnt);
	return 0;
}