PAT (Advanced Level) 1085. Perfect Sequence (25) 贪心算法

最新推荐文章于 2020-07-09 21:54:54 发布

tuzigg123

最新推荐文章于 2020-07-09 21:54:54 发布

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分类专栏： PAT(Advanced Level) 文章标签： PAT Advanced Level 1085 Perfect Sequence 贪心算法

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本文探讨了在给定整数序列和参数p的情况下，如何通过贪心算法找到最多能够形成'完美子序列'的整数数量。通过排序和动态调整来最大化选择范围，确保每个选择的整数乘以p不超过序列中的最大整数。

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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:

排序后使用贪心算法。

/*2015.7.30cyq*/
#include <iostream>
#include <vector>
#include <algorithm>
#include <fstream>
using namespace std;

//ifstream fin("case1.txt");
//#define cin fin

int main(){
	long long N,p;
	cin>>N>>p;
	vector<long long> ivec(N);
	for(int i=0;i<N;i++)
		cin>>ivec[i];

	sort(ivec.begin(),ivec.end());
	int low=0;
	int high=0;
	int len=0;
	while(1){
		while(high<N&&ivec[high]<=ivec[low]*p)
			high++;
		if(high-low>len)
			len=high-low;
		if(high==N)
			break;
		while(low<high&&ivec[low]*p<ivec[high])
			low++;
	}
	cout<<len;
	return 0;
}