Codeforces 158B

B. Taxi

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 s_i friends (1 ≤ s_i ≤ 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, ..., s_n (1 ≤ s_i ≤ 4). The integers are separated by a space, s_i 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 test(s)

Input

5 
1 2 4 3 3

Output

4

Input

8
2 3 4 4 2 1 3 1

Output

5

Note

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.

 

Problem analysis:

Just a greedy algorithm problem.And the number of the students in one group are only "1,2,3,4";

If we want the minimum number of taxis what we need.We just need let every taxi take the students as more as possible.

So if the group have 4 person,of course we need a car.

If the group have 3 person and the best way is to find another group noly have one person.

If the group have 2 person or 1 person, and that's easy.

Code view:

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    int num[5];
    int sum;
    int n;
    int i;
    while(cin>>n)
    {
        memset(num,0,sizeof(num));
        int a;
        for(i=1;i<=n;i++)
        {
             cin>>a;
             num[a]++;
        }
        sum=0;
        sum+=num[4];
        sum+=num[3];
        if(num[3]>=num[1])
            num[1]=0;
        else
            num[1]-=num[3];
        if(num[2]%2)
        {
            if(num[1]>1)
                num[1]-=2;
            else if(num[1]>0)
                num[1]--;
            sum+=1;
            sum+=num[2]/2;
        }
        else
        {
            sum+=num[2]/2;
        }
        if(num[1]%4)
            sum+=1;
        sum+=num[1]/4;
        cout<<sum<<endl;
    }
    return 0;
}