平方矩阵（）

平方矩阵1

平方矩阵2

蛇形矩阵

曼哈顿距离 

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>


using namespace std;

const int N = 110;

int n;
int a[N][N];

int main()
{
    while(cin >> n, n)
    {
        for (int i = 0; i < n; i ++ )
            for (int j = 0; j < n; j ++ )
            {
                if (n % 2)
                    a[i][j] = (n + 1) / 2 - max(abs(i - n / 2), abs(j - n / 2));
                else 
                    a[i][j] = (n - 1) / 2.0 - max(abs((n - 1) / 2.0- i), abs((n - 1) / 2.0 - j)) + 1;
            }

        for (int i = 0; i < n; i ++ )
        {
            for (int j = 0; j < n; j ++ ) 
                cout << a[i][j] << ' ';
            cout << endl;
        }

        cout << endl;
    }



    return 0;
}
2 2 2 2 2
2 1 1 1 2
2 1 0 1 2
2 1 1 1 2
2 2 2 2 2

 

 

从题意可以看出规律，每个点都是，该点到上下左右四条边的最小值 

while True:
    n = int(input())
    if not(n): break
    for i in range(n):
        for j in range(n):
            print(min(j, n - j - 1, i, n - i - 1) + 1, end = ' ')
        print()
    print()

1

1 1
1 1

1 1 1
1 2 1
1 1 1

1 1 1 1
1 2 2 1
1 2 2 1
1 1 1 1

1 1 1 1 1
1 2 2 2 1
1 2 3 2 1
1 2 2 2 1
1 1 1 1 1

蛇形矩阵 

n,m=map(int,input().split())

res=[ [0 for j in range(m)] for i in range(n)]#建立一个n行m列的二位矩阵

dx=[-1,0,1,0]#偏移量
dy=[0,1,0,-1]

x,y,d=0,0,1

for i in range(n*m):
    res[x][y]=i+1
    a=x+dx[d]
    b=y+dy[d]
    
    if a<0 or a>=n or b<0 or b>=m or res[a][b]>0:#遇到需要掉转方向的就需要转
        d=(d+1)%4
        a=x+dx[d]
        b=y+dy[d] 
        
    x,y=a,b

for i in res:
    for j in i:
        print(j,end=" ")
    print()