PAT 甲级 1085 Perfect Sequence

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本文详细解析了1085 PerfectSequence题目，介绍了两种解题方法：二分查找与双指针法。通过实例展示了如何利用这两种方法找到序列中形成完美子序列的最大数量元素。

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1085 Perfect Sequence （25 point(s)）

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:

经验总结：

这一题，两种方法，二分查找，或者用二指针，难度不大，先排序，再依次查找即可~

AC代码

二分查找：

#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=100010;
int a[maxn],n,p;
int main()
{
	scanf("%d%d",&n,&p);
	for(int i=0;i<n;++i)
		scanf("%d",&a[i]);
	sort(a,a+n);
	int len=-1;
	for(int i=0;i<n;++i)
	{
		int *x=upper_bound(a+i+1,a+n,(long long)a[i]*p);
		len=len>x-a-i?len:x-a-i;
	}
	printf("%d\n",len);
	return 0;
}

二指针：

#include <cstdio>
#include <algorithm>
using namespace std;
const int maxn=100010;
int a[maxn],n,p;
int main()
{
	scanf("%d%d",&n,&p);
	for(int i=0;i<n;++i)
		scanf("%d",&a[i]);
	sort(a,a+n);
	int len=1,i=0,j=0;
	while(i<n&&j<n)
	{
		while(j<n&&a[j]<=(long long)a[i]*p)
		{
			len=len>j-i+1?len:j-i+1;
			++j;
		}
		++i;
	}
	printf("%d\n",len);
	return 0;
}