pat 1085 Perfect Sequence

最新推荐文章于 2020-01-16 18:32:13 发布

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本文介绍了一个算法问题，即从给定的整数序列中找出尽可能多的数字形成一个完美子序列。其中，完美子序列定义为最大值与最小值之比不超过给定参数p的子序列。文章详细描述了输入输出规范，并提供了一个使用二分查找实现的示例代码。

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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:

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

vector<long long> vec;
long long binarySearch(long long istart, long long iend, long long num)
{
	while(istart <= iend) {
		long long mid = (istart + iend) / 2;

		if(vec[mid] == num) return mid;

		else if(vec[mid] > num) iend = mid-1;

		else istart = mid+1;
	}

	return istart;
}
int main() {
	long long n, p;

	scanf("%lld %lld", &n, &p);

	for (long long i = 0; i < n; ++i)
	{
		long long x;
		scanf("%lld", &x);
		vec.push_back(x);
	}

	sort(vec.begin(), vec.end());
	long long ans = 0;

	for (long long i = 0; i < n; ++i)
	{
		long long x = p * vec[i];
		long long pos = binarySearch(i+1, vec.size()-1, x);
		long long tmp = pos - i;
		if(pos < n && vec[pos] == x) tmp += 1; //如果x在数组内,加x加入

		if(tmp > ans) ans = tmp;
	}

	printf("%lld\n", ans);
	return 0;
}