1085. Perfect Sequence (25)

自己想的比较好的一个算法，时间大大节省

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:

8

来源： <http://www.patest.cn/contests/pat-a-practise/1085>

#include <iostream>
#include <stdio.h>
#include <vector>
#include <algorithm>
#pragma warning(disable:4996)

using namespace std;

vector<long long int> num;

int main(void) {
	int n, p, temp;
	cin >> n >> p;

	while (n--)
	{
		scanf("%d", &temp);
		num.push_back(temp);
	}

	sort(num.begin(), num.end());
	int count = 0, max = 0;

	n = num.size();
	for (int i = 0; i < n; i++) {
		count=0;
		if (n - max <= i) {
			break;
		}
		if (num[i + max] <= num[i] * p) {
			for (int j = i; j < n; j++) {
				if (num[j] <= num[i] * p) {
					count++;
				}
				else
					break;
			}
		}
		if (count > max)
			max = count;
	}

	cout << max;
	return 0;
}

来自为知笔记(Wiz)