https://ac.nowcoder.com/acm/contest/317/B

题解：贪心

先排4 0组合，如果一个有余与2组合，最后有余的排

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
ll t,n,m,k,q;
ll ans,cnt,flag,temp;
ll a[5];
char str;
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //scanf("%d",&n);
    //scanf("%d",&t);
    //while(t--){}
    scanf("%lld",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld",&k);
        a[k]++;
    }
    int pos=0;
    for(int i=1;i<=n;i++){
        if(pos==0){
            if(a[4]>0){
                ans+=16;
                pos=4;
                a[4]--;
            }else if(a[2]>0) {
                ans+=4;
                pos=2;
                a[2]--;

            }else{
                ans+=0;
                pos=0;
                a[0]--;
            }
        }else if(pos==2){
            if(a[0]>0){
                ans+=4;
                pos=0;
                a[0]--;
            }else if(a[4]>0) {
                ans+=4;
                pos=4;
                a[4]--;

            }else{
                ans+=0;
                pos=2;
                a[2]--;
            }


        }else{
            if(a[0]>0){
                ans+=16;
                pos=0;
                a[0]--;
            }else if(a[2]>0) {
                ans+=4;
                pos=2;
                a[2]--;
            }else{
                ans+=0;
                pos=4;
                a[4]--;
            }
        }

    }
    printf("%lld\n",ans);
    //cout << "Hello world!" << endl;
    return 0;
}