CodeForces 372A Counting Kangaroos is Fun （二分）

M - Counting Kangaroos is Fun

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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 s_i — the size of the i-th kangaroo (1 ≤ s_i ≤ 10⁵).

Output

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

Sample Input

Input

8
2
5
7
6
9
8
4
2

Output

5

Input

8
9
1
6
2
6
5
8
3

Output

5

题意：每个袋鼠可以把小于它体重一半的袋鼠，放进口袋，并且只能放一只，被放进口袋的袋鼠是看不见的，

给出N个袋鼠的体重，问最少有几个袋鼠一定被看见。

思路:先把N只袋鼠体重从小到大进行排序，从n/2到N进行二分，查找最小并且大于等于第一只袋鼠重量二倍的袋鼠

然后从此只袋鼠开始向后依次查找符合条件的（注意：被放进口袋的袋鼠口袋里不能再放）

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
int a[500010];
int main()
{
	int n,i,j,k,l,m;
	while(scanf("%d",&n)!=EOF)
	{
		for(i=0;i<n;i++)
		scanf("%d",&a[i]);
		sort(a,a+n);
		int l=n/2;
		int r=n-1;
		int ans;
		while(l<=r)
		{
			int mid=(l+r)/2;
			if(a[mid]>=2*a[0])
			{
				ans=mid;
				r=mid-1;
			}
			else
			l=mid+1;
		}
		k=0;
		//printf("%d\n",ans);
		int count=0;
		for(i=ans;i<n;i++)
		{
			if(k==ans)
			break;
			if(a[i]>=2*a[k])
			{
				count++;
				k++;
			}
		}
		printf("%d\n",n-count);
	}
	return 0;
}