Codeforces 373C Counting Kangaroos is Fun【贪心】

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

825769842

output

5

input

891626583

output

5

题目大意：

给出N个袋鼠，每个袋鼠的容量已知 ，如果一只袋鼠的容量是另一只袋鼠容量的二倍还要大的话，那么这只袋鼠就可以容纳另一只袋鼠，但是一只袋鼠只能容纳一只袋鼠，问最后露在外边的袋鼠的数量。

思路：

因为一只袋鼠只能容纳一只袋鼠，那么我们不妨将袋鼠按照容量从小到大排序之后，将其分成两部分，一部分是【1，n/2】，另一部分是【n/2+1，n】，那么对应我们用大的袋鼠去贪心的装小袋鼠即可。

Ac代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int a[1500050];int sum[1500050];int have[1500050];int main(){    int n;    while(~scanf("%d",&n))    {        int ans=0;        int now=0;        memset(have,0,sizeof(have));        memset(sum,0,sizeof(sum));        for(int i=1;i<=n;i++)scanf("%d",&a[i]);        sort(a+1,a+1+n);        int cnt=n;        for(int i=n/2;i>=1;i--)        {            if(a[i]*2<=a[cnt])ans++,cnt--;        }        printf("%d\n",n-ans);    }}