PAT (Advanced Level) Practice 1085 Perfect Sequence (25分)

1.题目

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 (≤10​5​​) is the number of integers in the sequence, and p (≤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

2.题目分析

PAT (Basic Level) Practice 1030 完美数列 (25分) (使用upper_bound进行二分查找!)

3.代码


#include<iostream>
#include<algorithm>
using namespace std;
long long list[100001];
int main()
{
	long long n, p;
	cin >> n >> p;
	for (int i = 0; i < n; i++)
		cin >> list[i];
	sort(list, list + n);
	long long maxa = 0;
	for (int i = 0; i < n; i++)
	{
		int index = upper_bound(list+i, list+n, list[i] * p)-list;
		if (index == n)index--;
        while(list[index]>list[i]*p)index--;
		maxa =(maxa> index - i+1)?maxa:(index-i+1);
	}
	cout << maxa;
}

 

posted @ 2020-04-15 10:29  Jason66661010  阅读(89)  评论(0编辑  收藏  举报