python递归有上限，so

import sys

sys.setrecursionlimit(3000)

import sys
sys.setrecursionlimit(3000)
n=int(input())
f=[[0 for i in range(100)]for j in range(100)]
def out(x,y,cnt):
    cnt+=1
    if x>n or y>n or x<1 or y<1 or cnt>n*n:
        return
    f[x][y]=cnt
    if x==1 and y!=n:
        out(n,y+1,cnt)
    elif y==n and x!=1:
        out(x-1,1,cnt)
    elif x==1 and y==n:
        out(x+1,y,cnt)
    elif x!=1 and y!=n:
        if f[x-1][y+1]==0:
            out(x-1,y+1,cnt)
        else:   
            out(x+1,y,cnt)
f[1][n//2+1]=1
out(1,(n+1)//2,0)
for i in range(1,n+1):
    for j in range(1,n+1):
        print(f[i][j],end=' ')
    print('')

n=int(input())
a=[[0 for i in range(n+10)] for j in range(n+10)]
#开二维数组
posx=[0]*40*40
posy=[0]*40*40
#posx，posy记录第i个数的x，y坐标
a[1][n//2+1]=1
posx[1]=1
posy[1]=n//2+1
for i in range (2,n*n+1):
    if posx[i-1]==1 and posy[i-1]!=n:
        a[n][posy[i-1]+1]=i
        posx[i]=n
        posy[i]=posy[i-1]+1
    elif posx[i-1]!=1 and posy[i-1]==n:
        a[posx[i-1]-1][1]=i
        posx[i]=posx[i-1]-1
        posy[i]=1
    elif posx[i-1]==1 and posy[i-1]==n:
        a[posx[i-1]+1][posy[i-1]]=i
        posx[i]=posx[i-1]+1
        posy[i]=posy[i-1]
    elif posx[i-1]!=1 and posy[i-1]!=n:
        if a[posx[i-1]-1][posy[i-1]+1]==0:
            a[posx[i-1]-1][posy[i-1]+1]=i
            posx[i]=posx[i-1]-1
            posy[i]=posy[i-1]+1
        else:
            a[posx[i-1]+1][posy[i-1]]=i
            posx[i]=posx[i-1]+1
            posy[i]=posy[i-1]
#简单易懂更符合新高二宝宝体质的for判断
#（虽然很丑
for i in range(1,n+1):
    for j in range(1,n+1):
        print(a[i][j],end=' ')
    print()

#include<bits/stdc++.h>
using namespace std;
const int N = 50;
int a[N][N],n,now;
void dfs(int x,int y){
    if(now==n*n+1)return;
    a[x][y]=++now;
    if(x==1 && y!=n) dfs(n,y+1);
    if(y==n && x!=1) dfs(x-1,1);
    if(x==1 && y==n) dfs(2,y);
    if(x!=1 && y!=n){
        if(a[x-1][y+1])dfs(x+1,y);
        else dfs(x-1,y+1);
    }
}
int main(){
    cin>>n;
    dfs(1,(n+1)/2);
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++)cout<<a[i][j]<<' ';
        puts("");
    }
    return 0;
}

#include <bits/stdc++.h>
using namespace std;

const int N = 40;

int m[N][N],n,t;

void dfs(int x,int y)
{
	if(t == n*n+1) return ;
	t++;
	m[x][y] = t;
	if(x == 1 && y!=n) dfs(n,y+1);
	else if(x!=1 && y == n) dfs(x-1,1);
	else if(x == 1 && y == n) dfs(x+1,y);
	else if(x!=1 && y!=n)
	{
		if(m[x-1][y+1]) dfs(x+1,y);
		else dfs(x-1,y+1);
	}
}

int main()
{
	cin >> n;
	dfs(1,(n+1)/2);
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
			cout << m[i][j] << " ";
		cout << endl;
	}
	return 0;
}

根据题意模拟即可，while判断

n=int(input())
a=[[0 for i in range(100)]for j in range(100)]
prei=1
prej=n//2+1
posj=0
posi=0
a[1][n//2+1]=1
k=2
while(k<=n*n):
   if(prei==1 and prej!=n):
       posi=n
       posj=prej+1
   elif(prei!=1 and prej==n):
       posi=prei-1
       posj=1
   elif(prei==1 and prej==n):
       posi=prei+1
       posj=prej
   elif(prei!=1 and prej!=n):
       if(a[prei-1][prej+1]==0):
           posi=prei-1
           posj=prej+1
       else:
           posj=prej
           posi=prei+1
   prei=posi
   prej=posj
   a[posi][posj]=k
   k+=1
for i in range(1,n+1):
    for j in range(1,n+1):
        print(a[i][j],end=" ")
    print()

python版本的深搜！

sys.setrecursionlimit(3000)
def dfs(x,y,q):
    if q==n*n+1:
        return 0
    a[x][y]=q
    q=q+1
    if x==1 and y!=n:
        dfs(n,y+1,q)
    if x!=1 and y==n:
        dfs(x-1,1,q)
    if x==1 and y==n:
        dfs(x+1,y,q)
    if x!=1 and y!=n:
        if a[x-1][y+1]==0:
            dfs(x-1,y+1,q)
        else:
            dfs(x+1,y,q)
    return 0

n=int(input())
a=[[0 for i in range(n+2)] for j in range(n+2)]
q=1
p=(n+1)//2
dfs(1,p,q)
for i in range(1,n+1):
    for j in range(1,n+1):
        print(a[i][j],end=' ')
    print()