codeforces839B(88/600)

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Daenerys Targaryen has an army consisting of k groups of soldiers, the i-th group contains ai soldiers. She wants to bring her army to the other side of the sea to get the Iron Throne. She has recently bought an airplane to carry her army through the sea. The airplane has n rows, each of them has 8 seats. We call two seats neighbor, if they are in the same row and in seats {1, 2}, {3, 4}, {4, 5}, {5, 6} or {7, 8}.

A row in the airplane
Daenerys Targaryen wants to place her army in the plane so that there are no two soldiers from different groups sitting on neighboring seats.

Your task is to determine if there is a possible arranging of her army in the airplane such that the condition above is satisfied.

Input
The first line contains two integers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 100) — the number of rows and the number of groups of soldiers, respectively.

The second line contains k integers a1, a2, a3, …, ak (1 ≤ ai ≤ 10000), where ai denotes the number of soldiers in the i-th group.

It is guaranteed that a1 + a2 + … + ak ≤ 8·n.

Output
If we can place the soldiers in the airplane print “YES” (without quotes). Otherwise print “NO” (without quotes).

You can choose the case (lower or upper) for each letter arbitrary.

Examples
input
2 2
5 8
output
YES
input
1 2
7 1
output
NO
input
1 2
4 4
output
YES
input
1 4
2 2 1 2
output
YES
Note
In the first sample, Daenerys can place the soldiers like in the figure below:

In the second sample, there is no way to place the soldiers in the plane since the second group soldier will always have a seat neighboring to someone from the first group.

In the third example Daenerys can place the first group on seats (1, 2, 7, 8), and the second group an all the remaining seats.

In the fourth example she can place the first two groups on seats (1, 2) and (7, 8), the third group on seats (3), and the fourth group on seats (5, 6).

卧槽大模拟…
题解都在注释里了…

讲道理，模拟题写注释效率真的高

#include<bits/stdc++.h>using namespace std;int tu[101];int main(){    int n,m;    cin>>n>>m;    for(int a=1;a<=m;a++)scanf("%d",&tu[a]);    int si=n,er=2*n;    for(int a=1;a<=n&&si>0;a++)    {        int zs=tu[a]/4;        if(zs<=si)si-=zs,tu[a]%=4;        else tu[a]-=si*4,si=0;    }    if(si)    {//全部都剩下了不到4的         for(int a=1;a<=m&&si>0;a++)        {//尽量拿走3的             if(tu[a]==3)            {                tu[a]=0;                si--;             }        }        if(si)        {            //3的放完了还有四连坐接着放1和2            int yiyi=0,erer=0;            for(int a=1;a<=m;a++)            {                if(tu[a]==1)yiyi++;                if(tu[a]==2)erer++;            }            int yierxiao=min(yiyi,erer);            if(yierxiao<=si)            {                //如果还有四连坐能放下其中一部分...                 yiyi-=yierxiao;                erer-=yierxiao;                si-=yierxiao;                if(yiyi)                {                    //如果剩下1的                     if(si*2+er>=yiyi)cout<<"YES";                    else cout<<"NO";                    return 0;                }                if(erer)                {                    //如果剩下2的                     if(er>=erer)                    {                        cout<<"YES";                        return 0;                    }                    erer-=er;                    int zs2=erer/3;                    if(si<zs2*2)                    {                        cout<<"NO";                        return 0;                    }                    si-=zs2*2;                    erer-=zs2*3;                    if(si>=erer)                    {                        cout<<"YES";                    }                    else cout<<"NO";                    return 0;                }            }            else            {                //如果放不下了..                yiyi-=si;                erer-=si;                si=0;                if(er>=yiyi+erer)cout<<"YES"<<endl;                else cout<<"NO";                return 0;            }        }        else        {            //如果三的都放不进四连坐里...             int yiyi=0,erer=0;            for(int a=1;a<=m;a++)            {                if(tu[a]==1)yiyi++;                else if(tu[a]==2)erer++;                else if(tu[a]==3)yiyi++,erer++;            }            if(yiyi+erer<=er)cout<<"YES";            else cout<<"NO";            return 0;        }    }    else    {        //连能%4的都装不下...        int yiyi=0,erer=0;        for(int a=1;a<=m;a++)        {            if(tu[a]%2)yiyi++,erer+=tu[a]/2;            else erer+=tu[a]/2;        }         if(yiyi+erer<=er)cout<<"YES";        else cout<<"NO";        return 0;    }}