#1085. Perfect Sequence【双指针 + 二分】

原题链接

Problem Description：

Given a sequence of positive integers and another positive integer $p$. The sequence is said to be a perfect sequence if $M\le m\times p$ 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 $N$ ($\le 10^5$) is the number of integers in the sequence, and $p$ ($\le 10^9$) is the parameter. In the second line there are $N$ positive integers, each is no greater than $10^9$.

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:

8

Problem Analysis:

算法 1（双指针 + 二分）：

先将所给序列从大到小排序，然后枚举最小值 $m$，设定一个全局最大值 maxlen，然后二分出满足要求的最大的 $M$，不断更新 maxlen 即可，时间复杂度为 $O(n\log n)$。

算法 2（双指针）：

由于可以证明在双指针过程中，当前这一轮的最大值 $M$ 向右移的过程中，最小值 $m$ 一定不会向左移动或者不变，并且一定也会向右移，故可以直接进行双指针，并且每一个元素最多只会被遍历一次，这种做法时间复杂度是线性的。

证明过程：

设当前枚举的最小值为 $m$，最大值为 $M$，向右移动后最大值为 $M^{\prime }$，$m$ 移动后为 $m^{\prime }$，其中 $m^{\prime }\le m$，而又 $\because$ $M\le m\times p,M^{\prime }>m\times p\ge m^{\prime }\times p$，矛盾。所以 $M$ 在向右移动的过程中，$m$ 也会顺次向右移动。

Code

双指针 + 二分

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 1e5 + 10;

int n, p;
int a[N];

bool check(int min_id, int max_id)
{
    
    
    if (a[max_id] <= (LL)a[min_id] * p) return true;
    return false;
}

int main()
{
    
    
    cin >> n >> p;
    for (int i = 0; i < n; i ++ ) 
        scanf("%d", &a[i]);
    
    sort(a, a + n);
    // 1 2 3 4 5 6 7 8 9 20
    // maxa <= mina * p
    
    int maxlen = -1;
    for (int i = 0; i < n; i ++ ) // 枚举最小数，即起始位置
    {
    
    
        int l = i + 1, r = n - 1; // 找到第一个不满足这个条件的边界
        while (l < r)
        {
    
    
            int mid = l + r + 1 >> 1;
            if (check(i, mid)) l = mid;
            else r = mid - 1;
        }
        int len = r - i + 1;
        maxlen = max(maxlen, len);
    }
    cout << maxlen << endl;
    return 0;
}

双指针

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1e5 + 10;

int n, p;
int a[N];

int main()
{
    
    
    cin >> n >> p;
    for (int i = 0; i < n; i ++ ) scanf("%d", &a[i]);
    
    sort(a, a + n);
    
    int res = 0;
    for (int i = 0, j = 0; i < n; i ++ )
    {
    
    
        while ((long long)a[j] * p < a[i]) j ++ ; // a[j] 是 m，a[i]是 M
        res = max(res, i - j + 1);
    }
    cout << res << endl;
    return 0;
}