1085 Perfect Sequence （25 分）

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 (≤10⁹) 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

題目大意

給定一個序列A和引數p，若序列A中最大值為M，最小值為m，滿足M<=m*p，則稱序列A為perfect sequence；依據這個定義，題目要求在給定的序列中找出儘可能多的元素組成一個新的序列，使得這個序列為perfect sequence。

分析

尋找序列極值問題，若對演算法的複雜度無苛刻要求，可先對序列排序，定義當前最大寬度maxwidth，遍歷序列，遍歷過程中以當前maxwidth為起始寬度尋找比maxwidth更大寬度的序列。另外，題目中序列元素和引數p的範圍均為0~10⁹,兩數想成數的範圍將可能超過10⁹，因此將數的型別定義為long int

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> v;
int main(){
	long long n,p;
	cin>>n>>p;
	v.resize(n);
	for(int i=0;i<n;i++) cin>>v[i];
	sort(v.begin(), v.end());
	int maxwidth=1;
	for(int i=0;i<n;i++){
		while(i+maxwidth-1<n && v[i+maxwidth-1]<=p*v[i]) maxwidth++;
	}
	cout<<maxwidth-1;
}