ACM-ICPC 2017 Asia Nanning-J：Rearrangement（思维）

In a two dimensional array of integers of size $2\times n$, is it possible to rearrange integers so that the sum of two adjacent elements (which are adjacent in a common row or a common column) is never divisible by three?

Input

The input has several test cases and the first line contains an integer $t(1\le t\le 200)$ which is the number of test cases.

In each case, the first line contains an integers $n(1\le n\le 10000)$ indicating the number of columns in the array. The second line contains the elements of the array in the first row separated by single spaces. The third line contains the elements of the array in the second row separated by single spaces. The elements will be positive integers less then $1000000$.

Output

For each test case, output “YES” in a single line if any valid rearrangement exists, or “NO” if not.

样例输入

6
3
3 6 9
1 4 7
3
3 6 9
1 3 8
5
1 2 3 4 5
6 7 8 9 10
10
1 1 1 1 1 1 1 1 1 1
2 3 2 3 2 3 2 3 2 3  
2
3 1
2 3
2
3 1
1 2

样例输出

YES
NO
YES
YES
YES
NO

思路：可以把数分成三种：即除3的余数分别为0，1，2。那么余数为1，2的2种数不能相邻，余数为0，0的2种数不能相邻。那么可以考虑余数为1的数放在最左边，余数为2的数放在最右边，余数为0的数放在中间。如图：

1	1	1	0	2	2	2	2
1	1	1	1	0	2	2	2

 不过还有些细节要处理。

#include
using namespace std;
int a[4];
int main()
{
    int T;cin>>T;
    while(T--)
    {
        int n;
        scanf("%d",&n);
        memset(a,0,sizeof a);
        for(int i=1;i<=2*n;i++)
        {
            int x;
            scanf("%d",&x);
            a[x%3]++;
        }
        if(a[0]>n)puts("NO");//当0的个数大于n，必定有2个0相邻
        else
        {
            if(a[0]<=1&&a[1]&&a[2])puts("NO");//当0的个数小于等于1时，必定有1个1和1个2相邻
            else if(a[1]==0||a[2]==0)puts("YES");
            else if(a[0]>=2&&a[1]&&a[2])
            {
                if(a[0]==2&&a[1]%2==0&&a[2]%2==0)puts("NO");//当只有2个0且1和2的个数均为偶数时，必定有1个1和1个2相邻
                else puts("YES");
            }
        }
    }
    return 0;
}