made in china

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int M=21;
int l[5]={0,-1,0,1,0};
int r[5]={0,0,1,0,-1};
int a[M][M],b[M*M][3];// 螺旋矩阵的值、各值的坐标
int  c[M*M],n,k; //c记录“王”棋子遍历的轨迹
bool d[M*M],flag=false; //记录矩阵中各个位置是否遍历

void dfs(int t,int x1, int y1) 
//从第t步开始，当前位置在(x1,y1)时的遍历路径
{
  int  i,x,y;
  if (flag) return;
  if (t>n*n)
  {
      cout<<c[1];
      for (i=2; i<=n*n; i++) cout<<' '<<c[i];
      cout<<endl;
      flag=true; return;
  }
  for (i=1; i<=4; i++)
  {
    x=x1+l[i];  y=y1+r[i];
    if ((x>=1)&&(x<=n)&&(y>=1)&&(y<=n)&&(d[a[x][y]]))
    {
      d[a[x][y]]=!d[a[x][y]];
      c[t]=a[x][y];
      dfs(t+1,x,y);
      c[t]=0;
      d[a[x][y]]=!d[a[x][y]];
    }
  }
}

void out2()  //处理奇数情况，根据情况搜索
{
  memset(c,0,sizeof(c));
  memset(d,true,sizeof(d));
  if ((b[k][1]+b[k][2])%2==1) cout<<"No Path"<<endl;
  else
  {
    c[1]=k;
    d[k]=!d[k];
    dfs(2,b[k][1],b[k][2]);
  }
}

void out1()
//处理n为偶数情况，按上图所示的方式产生回路
{
  int d[M*M],x,y,i,t,f,p,z;
  for (i=1; i<=n; i++) d[i]=i;
  t=n*n/2+1;
  for (i=n; i>=1; i--)
  {
    d[t]=a[1][i]; t++;
  }
  t=n+1; z=n*n/2; x=n-1;  y=n;  f=-1;
  while (t<=z)
  {
    d[t]=a[x][y]; t++;
    if ((f==-1)&&(y>n/2+1)) y--;
    else if ((f==-1)&&(y==n/2+1)) {x--; f=1;}
    else if ((f==1) &&(y<n)) y++;
    else { f=-1; x--; }
  }
  t=n*n/2+n+1;  f=1; x=2;  y=1;
  while (t<=n*n)
  {
    d[t]=a[x][y]; t++;
    if ((f==1)&&(y<n/2)) y++;
    else if ((f==1)&&(y==n/2)) {f=-1; x++;}
    else if ((f==-1)&&(y>1)) y--;
    else { f=1;  x++;}
  }
  p=1;
  while (d[p]!=k) p++;
  for (i=p; i<=n*n; i++) cout<<d[i]<<' ';
  for (i=1; i<p; i++)
    if (i!=(p-1)) cout<<d[i]<<' ';
    else  cout<<d[p-1]<<endl;
}

void jz()
{
  int i,j,x,y,z,t,f;
  x=n;  y=1;  t=1;  f=1;  z=n*n;
  while (t<=z)
  {
    a[x][y]=t; b[t][1]=x; b[t][2]=y;
    t++;
    if ((f==1)&&(y<n)&&(a[x][y+1]==0)) y++;
    else if ((f==1)&&((y==n)||(a[x][y+1]>0))){f=2; x--;}
    else if ((f==2)&&(x>1)&&(a[x-1][y]==0)) x--;
    else if ((f==2)&&((x==1)||(a[x-1][y]>0))) {f=3; y--;}
    else if ((f==3)&&(y>1)&&(a[x][y-1]==0)) y--;
    else if ((f==3)&&((y==1)||(a[x][y-1]>0))) {f=4; x++;}
    else if ((f==4)&&(x<n)&&(a[x+1][y]==0)) x++;
    else { f=1; y++;}
  }
  for (i=1; i<=n; i++)
  {
    for (j=1; j<=n; j++)  printf("%4d",a[i][j]);
    printf("\n");
  }
}

int main()
{
  cin>>n>>k;  jz();
  if (n%2==0) out1();
  else out2();
  fclose(stdin); fclose(stdout);
  return 0;
}