【PAT】【Advanced Level】1085. Perfect Sequence (25)

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本文介绍了一个算法问题——寻找最长的完美子序列。通过输入一系列正整数和参数p，目标是找出满足最大值与最小值比不超过p的最大子序列长度。文章提供了具体的输入输出规格，并给出了一种使用排序加二分查找的解决方案。

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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:

原题链接：
https://www.patest.cn/contests/pat-a-practise/1085

https://www.nowcoder.com/pat/5/problem/4035

思路：

排序+二分查找

坑点：

PAT上最后一个数据点要用longlong

CODE：

#include<iostream>
#include<algorithm>
#include<cstdio>
#define N 100010
using namespace std;
long long  num[N];
int main()
{
    int n;
	long long p;
    scanf("%d %ld",&n,&p);
    for (int i=0;i<n;i++)
        scanf("%ld",&num[i]);
    sort(num,num+n);
    int mx=0;
    for (int i=0;i<n;i++)
    {
        if (i>=n-mx) break;
        int l=i;
        int r=n-1;
        int mid;
        while (l<=r)
        {
            mid=(l+r)/2;
            if (num[mid]>num[i]*p)
            {
                r=mid-1;   
            }
            else
            {
                l=mid+1;
            }
        }
        mx=max(mx,l-i);
    }
    cout<<mx;
    return 0;
}