CodeForces 373C Counting Kangaroos is Fun（袋鼠藏口袋，二分）

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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
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：

有 n 个袋鼠的口袋，袋鼠 A 的口袋 >= 袋鼠 B 口袋的 2 倍，后者就能够藏在前者口袋中。

问最后最多能剩下多少个能够看到的袋鼠

思路：

跟之前做的一串数字组合类似。。。

二分查找方法

代码：

#include<cstdio>
#include<algorithm>
#define MYDD 11030

using namespace std;

int num[50*MYDD];
bool cmp(int x,int y) {
	return x>y;
}

int main() {
	int n;
	scanf("%d",&n);
	for(int j=0; j<n; j++)
		scanf("%d",&num[j]);

	sort(num,num+n,cmp);

	int left=n/2,del=0;	
	//int v=1;
	for(int j=0; j<n; j++) {
		if(num[j]==0)
			continue;
		while(left<n) {//折半查找满足条件的值 
			if(num[left]!=0&&(2*num[left]<=num[j])) {
				num[j]=num[left]=0;//找到一组这个袋鼠能够待在另一个袋鼠口袋中 
				del++;
				left++;
				break;
			} else
				left++;
			//printf("*************%d**\n",v++);
		}

	}
	printf("%d\n",n-del);//剩余可见的袋鼠 
	return 0;
}

RYC学长的代码：

#include<iostream>  
#include<cstdio>  
#include<cstdlib>  
#include<cstring>  
#include<algorithm>  
#include<cmath>  
#include<queue>  
#include<stack>  
#include<list>  
#include<functional>  
using namespace std;  
const int maxn=1000010;  
int num[maxn],temp[maxn],n;  
bool judge(int mid){  
    if(mid<(n+1)/2)return false;  
    for(int i=mid+1;i<=n;++i){  
        if(num[i]*2>num[i-mid])return false;  
    }  
    return true;  
}  
bool cmp(int a,int b){  
    return a>b;  
}  
int main()  
{  
    scanf("%d",&n);  
    for(int i=1;i<=n;++i){  
        scanf("%d",&num[i]);  
    }  
    sort(num+1,num+n+1,cmp);  
    int left=1,right=n,ans;  
    while(left<=right){  
        int mid=(left+right)/2;  
        if(judge(mid)){  
            ans=mid;  
            right=mid-1;  
        }  
        else {  
            left=mid+1;  
        }  
    }  
    printf("%d\n",ans);  
    return 0;  
}