ZCMU 1424

最新推荐文章于 2022-06-22 21:57:16 发布

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最新推荐文章于 2022-06-22 21:57:16 发布

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本文介绍了一个算法问题，涉及一群孩子按小组庆祝生日，每组1到4人，他们需要乘坐出租车前往目的地。目标是最小化所需出租车数量，同时确保来自同一小组的孩子们能乘坐同一辆出租车。文章详细阐述了解决这一问题的算法实现，包括如何分配孩子到出租车中，以达到最少车辆数。

1424: Taxi

Description

After the lessons n groups of schoolchildren went outside and decided to visit Polycarpus to celebrate his birthday. We know that the i-th group consists of si friends (1 ≤ si ≤ 4), and they want to go to Polycarpus together. They decided to get there by taxi. Each car can carry at most four passengers. What minimum number of cars will the children need if all members of each group should ride in the same taxi (but one taxi can take more than one group)?

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of groups of schoolchildren. The second line contains a sequence of integers s1, s2, …, sn (1 ≤ si ≤ 4). The integers are separated by a space, si is the number of children in the i-th group.

Output

Print the single number — the minimum number of taxis necessary to drive all children to Polycarpus.

Sample Input

5
1 2 4 3 3
8
2 3 4 4 2 1 3 1

Sample Output

4
5

HINT

In the first test we can sort the children into four cars like this:

·the third group (consisting of four children),

·the fourth group (consisting of three children),

·the fifth group (consisting of three children),

·the first and the second group (consisting of one and two children, correspondingly).

There are other ways to sort the groups into four cars.

Code:

#include<stdio.h>
#include<string.h>
int main()
{
    int n;
    int a[100005],x[100005];
    while(scanf("%d",&n)!=EOF)
    {
        memset(x,0,sizeof(x));
        int sum=0;
        for(int i=0; i<n; i++)
        {
            scanf("%d",&a[i]);
            x[a[i]]++;
        }
        sum+=x[4];
        //printf("%d\n",sum);
        x[1]-=x[3];
        sum+=x[3];
        //printf("%d\n",sum);
        int m=x[2]/2;
        sum+=m;
        x[2]-=m*2;
        if(x[2]>0)
        {
            sum+=x[2];
            x[1]-=x[2]*2;
        }
        if(x[1]>0)
        {
            if(x[1]%4==0)sum+=x[1]/4;
            else sum+=x[1]/4+1;
        }
        printf("%d\n",sum);
    }
    return 0;
}