PAT (Advanced Level) Practise 1085 Perfect Sequence (25)

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最新推荐文章于 2017-10-05 18:07:03 发布

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本文探讨了在给定整数序列和参数p的情况下，如何找出最多可以形成完美子序列的整数数量。通过排序和双指针法，实现高效求解。

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1085. Perfect Sequence (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CAO, Peng

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:

搞两个指针循环一下就好了。

#include<cstdio>
#include<stack>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 10;
int n, p, a[maxn], ans;

int main()
{
	scanf("%d%d", &n, &p);
	for (int i = 0; i < n; i++) scanf("%d", &a[i]);
	sort(a, a + n);
	for (int i = 0, j = 0; i < n; i++)
	{
		while (j < n && (LL)a[j] <= (LL)a[i] * p) j++;
		ans = max(j - i, ans);
	}
	printf("%d\n", ans);
	return 0;
}