C. Magic Grid 构造矩阵

C. Magic Grid

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Let us define a magic grid to be a square matrix of integers of size n×n, satisfying the following conditions.

• All integers from exactly once.
• Bitwise XOR of all elements in a row or a column must be the same for each row and column.

You are given an integer n×n.

Input

The only line of input contains an integer 4.

Output

Print a magic grid, i.e. i-th row of the grid.

If there are multiple answers, print any. We can show that an answer always exists.

Examples

input

Copy

4

output

Copy

8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14

input

Copy

8

output

Copy

19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61

Note

In the first example, XOR of each row and each column is 13.

In the second example, XOR of each row and each column is 60.

 

 题意：构造一个这样的矩阵：第 i 行的异或和==第 i 列的异或和

 

题解：若已知一个满足题意的矩阵，对这个矩阵的每一个数加上一个4的倍数依然满足题意，又因为n是4的倍数，所以不断用这样的4*4矩阵凑成一个n*n的矩阵即可

 

#include<iostream>
#include<algorithm>
using namespace std;
int a[1024][1024];
int main()
{
    int n,cnt=0;
    cin>>n;
    for(int i=1;i<=n;i=i+4)
    {
        for(int j=1;j<=n;j=j+4)
        {
            for(int x=0;x<4;x++)
            {
                for(int y=0;y<4;y++)
                {
                    a[i+x][j+y]=cnt++;
                }
            }
        }
    }
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(j==1)
                cout<<a[i][j];
            else
                cout<<' '<<a[i][j];
        }
        cout<<endl;
    }
    return 0;
}