Codeforce Round #219 Div2 C

C. Counting Kangaroos is Fun

time limit per test 1 second

memory limit per test 256 megabytes

 这么简单的贪心都没想到！，难道就只知道首尾和相邻？这个是由于最多就n/2的数量，也尽量消耗n/2以内的大的，因为消耗了n/2以内的大的，即使用了最大的也没关系，如果这样了，n/2小的也有可能消耗，如果不这样，也就顶多用最大的消耗一个n/2稍微小的！

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 test(s)

Input

8 
2 5 7 6 9 8 4 2

Output

5

Input

8 
9 1 6 2 6 5 8 3

Output

5

#include <cstdio>
#include <vector>
#include <cmath>
#include <queue>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define INF 0x7fffffff
#define mod 1000000007
#define ll unsigned long long int
#define maxn 1005
int n, w, k, m, ans;
int dp[maxn][maxn];
int a[101000];
int main(){
    cin >> n;
    for (int i = 1; i <= n; i++)cin >> a[i];
    sort(a + 1, a + n + 1);
    int i = n,j = n / 2,ans=0;
    while (i>=n/2&&j>0){
        if (a[i] >= 2 * a[j])i--, j--, ans++;
        else j--;
    }
    cout << n-ans;
    return 0;
}