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 nrows, 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.
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.
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.
2 2 5 8
YES
1 2 7 1
NO
1 2 4 4
YES
1 4 2 2 1 2
YES
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).
思路：最简单直接的做法就是加上无数的特判，然后卡过。主要思路是注意特判时的优先级。在此，主要就是注意以下几组数据：
输入：
2 7
2 2 2 2 2
输出：
YES
输入：
1 4
2 2 1 2
输出：
YES
输入：
1 4
2 2 2 2
输出：
NO
AC代码：
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#define MAX 10005
using namespace std;
int main( )
{
int n,m;
while(cin >> n >> m){
int seat4 = n,seat2 = n*2,seat1 = 0;
int need_seat4,need_seat2;
for(int j = 0; j < m; j++){
int a;
cin >> a;
while(a > 0){
if(seat4 > 0 && a >= 3)
seat4--,a -= 4;
else if(seat2 > 0 && a >= 2)
seat2--,a -= 2;
else if(a == 1)
seat1--,a--;
else if(seat4 > 0)
seat4--,a -= 2,seat1++;
else seat1--,a--;
}
}
if(seat4 >= 0 && seat4*2 + seat2 + seat1 >= 0)
cout << "YES" << endl;
else
cout << "NO" << endl;
}
return 0;
}不过不用特判直接暴力的也有：
#include<iostream>
#include<cstring>
#define MAX 10005
using namespace std;
int num[MAX];
int c[5];
int main( )
{
int n,m;
while(cin >> n >> m){
c[2] = n*2;
c[4] = n;
int a;
memset(num,0,sizeof(num));
for(int j = 0; j < m; j++) cin >> a,num[a]++;
for(int j = MAX; j; j--){
while(num[j]--){
int k;
for(k = 4; k; k--){
if(c[k]){
int t = min(j,k);
num[j - t]++;
c[k]--;
if(k - t - 1 > 0) c[k - t - 1]++;
break;
}
}
if(!k){
cout << "NO" << endl;
goto skip;
}
}
}
cout << "YES" << endl;
skip:;
}
}#include<iostream>
int i,n,k,a,s,j;
main(){
for(std::cin>>n>>k;i<k;i++)std::cin>>a,a%2?a++,j++:0,s+=a;
std::cout<<(s>8*n||s==8*n&&k==4*n&&j<n?"NO":"YES");
}这个B题还是比较坑的，在终测结束后。8000+ 人只有200+人通过。只能说这个B有毒。