首页
学习
活动
专区
圈层
工具
发布
首页
学习
活动
专区
圈层
工具
MCP广场
社区首页 >专栏 >【CodeForces】373C - Counting Kangaroos is Fun(二分)

【CodeForces】373C - Counting Kangaroos is Fun(二分)

作者头像
FishWang
发布2025-08-26 20:34:18
发布2025-08-26 20:34:18
17900
代码可运行
举报
运行总次数:0
代码可运行

点击打开题目

C. Counting Kangaroos is Fun

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There are n kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.

Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.

The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.

Input

The first line contains a single integer — n (1 ≤ n ≤ 5·105). Each of the next n lines contains an integer si — the size of the i-th kangaroo (1 ≤ si ≤ 105).

Output

Output a single integer — the optimal number of visible kangaroos.

Examples

input

代码语言:javascript
代码运行次数:0
运行
复制
8
2
5
7
6
9
8
4
2

output

代码语言:javascript
代码运行次数:0
运行
复制
5

input

代码语言:javascript
代码运行次数:0
运行
复制
8
9
1
6
2
6
5
8
3

output

代码语言:javascript
代码运行次数:0
运行
复制
5

在 0 和 n/2 之间对能干掉的人数二分。

代码如下:

代码语言:javascript
代码运行次数:0
运行
复制
#include <cstdio>
#include <algorithm>
using namespace std;
int num[500000+11];
int n;
bool check(int x)
{
	int pos = x - 1;
	for (int i = 0 ; i < x ; i++)
	{
		pos = lower_bound (num + pos + 1 , num + n , 2*num[i]) - num;
		if (pos == n)
			return false;
	}
	return true;
}
int main()
{
	while (~scanf ("%d",&n))
	{
		for (int i = 0 ; i < n ; i++)
			scanf ("%d",&num[i]);
		sort(num , num + n);
		int l = 0;
		int r = n / 2 + 1;
		int mid;
		while (r >= l)
		{
			mid = (l + r) >> 1;
			if (check(mid))
				l = mid + 1;
			else
				r = mid - 1;
		}
		printf ("%d\n",n-r);
	}
	return 0;
}
本文参与 腾讯云自媒体同步曝光计划,分享自作者个人站点/博客。
原始发表:2016-07-26,如有侵权请联系 cloudcommunity@tencent.com 删除

本文分享自 作者个人站点/博客 前往查看

如有侵权,请联系 cloudcommunity@tencent.com 删除。

本文参与 腾讯云自媒体同步曝光计划  ,欢迎热爱写作的你一起参与!

评论
登录后参与评论
0 条评论
热度
最新
推荐阅读
领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档