https://codeforces.com/problemset/problem/1208/C
Let us define a magic grid to be a square matrix of integers of size n×nn×n, satisfying the following conditions.
- All integers from 00 to (n2−1)(n2−1) inclusive appear in the matrix 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 nn which is a multiple of 44. Construct a magic grid of size n×nn×n.
Input
The only line of input contains an integer nn (4≤n≤10004≤n≤1000). It is guaranteed that nn is a multiple of 44.
Output
Print a magic grid, i.e. nn lines, the ii-th of which contains nn space-separated integers, representing the ii-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 1313.
In the second example, XOR of each row and each column is 6060.
好嘛最开始猜就是
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
这样的矩阵，然后没去算.至于怎么发现这个矩阵的....构造题很玄学。
考虑到题目只出4的倍数，猜测是一种暗示，暗示分成4x4的块，后面的数在前面的数上加16,32....
因为现在这个1号矩阵为位是2^0,2^1,2^2,2^3。隔壁2号矩阵加上16的话，在2^4加，而偶次2^4进行异或答案还是0
这道题获得的小规律：以偶数开始连续两个数如4^5=1;以偶数开始连续四个数如2^3^4^5=0；
要真是赛场上的话这种没啥特别性的构造还是往特殊了去构造
#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<cstdio>
#include<algorithm>
#define debug(a) cout<<#a<<"="<<a<<endl;
using namespace std;
const int maxn=1e3+100;
typedef long long LL;
LL a[maxn][maxn];
LL cnt=0;
void solve(LL x,LL y)
{
for(LL i=0;i<4;i++)
for(LL j=0;j<4;j++){
a[x+i][y+j]=cnt++;
}
}
int main(void)
{
cin.tie(0);std::ios::sync_with_stdio(false);
LL n;
cin>>n;
for(LL i=1;i<=n;i+=4)
for(LL j=1;j<=n;j+=4)
{
solve(i,j);
}
for(LL i=1;i<=n;i++)
{
for(LL j=1;j<=n;j++)
{
cout<<a[i][j]<<" ";
}
cout<<endl;
}
return 0;
}