839B Game of the Rows（水，思维，贪心）

B. Game of the Rows

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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 hasn 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.

题目思路：
比赛的时候虽然想到了策略，但是还是没有想清楚细节，导致实现起来非常麻烦，想清楚后其实就是道水题，以后还是静下来想清楚再写，对于这道题，我们可以想到先填四个的，然后填两个的，再填一个的，也就是把每一个数拆成4+2+1的形式，唯一需要注意的就是当填完四的填2的时候，要加上剩余可以填4的数量，因为不能连续填两个二，另外在最后填1的时候要加上剩下的两个的

ac代码：

#include<cstdio>#include<algorithm>#include<cstring>#include<cstring>#include<iostream>#include<sstream>#include<cmath>#include<vector>#define LL long long#define INF 0x3f3f3f3f#define eps 1e-6#include<deque>using namespace std;const int maxn = 200000+50;int n,k;int a[105];int main(){    while(~scanf("%d%d",&n,&k)){        for(int i = 0;i<k;i++){            scanf("%d",&a[i]);        }        int sum1 = n;        int sum2 = 2*n;        for(int i = 0;i<k;i++){            int d = min(sum1,a[i]/4);            sum1 -= d;            a[i] -= d*4;        }        sum2+=sum1;        for(int i = 0;i<k;i++){            int d = min(sum2,a[i]/2);            sum2 -= d;            a[i] -= d*2;          //  cout<<a[i]<<endl;        }        int tmp  = sum2+sum1;        for(int i = 0;i<k;i++){            //    cout<<a[i]<<endl;            tmp -= a[i];        }        if(tmp>=0){            puts("YES");        }        else            puts("NO");    }}