Codeforces-Counting Kangaroos is Fun【贪心】【二分】

最新推荐文章于 2020-07-08 11:59:30 发布

Somethingwll

最新推荐文章于 2020-07-08 11:59:30 发布

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分类专栏：大菜鸟小历程贪心二分查找

本文链接：https://blog.youkuaiyun.com/wang907553141/article/details/52049846

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本文探讨了一种有趣的算法问题——如何将多个不同大小的袋鼠按照特定规则装入其他袋鼠的口袋中，并求出最少可见袋鼠的数量。通过两种不同的代码实现方式，展示了高效的解决方案。

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Counting Kangaroos is Fun

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

Submit Status

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

Output

Input

Output

题意：有N个袋鼠每个袋鼠的口袋里可以放一只体重小于其体重的小于它二分之一重量的袋鼠现在将一些袋鼠放进其它的袋鼠口袋里问最多能见到多少袋鼠。

代码1：

#include<cstdio>
#include<algorithm>
#define Q 500000+10
using namespace std;
int n,a[Q];
int main()
{
	while(~scanf("%d",&n))
	{
		for(int i=0;i<n;i++)
			scanf("%d",&a[i]);
		sort(a,a+n);
		int ans=n-1;
		for(int i=n/2-1;i>=0;i--)
		{
			if(a[i]*2<=a[ans])	ans--;
		}
		printf("%d\n",ans+1);
	}
	return 0;
}

代码2：

#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;  
}