风之语故乡

这是我自己写的，很粗陋了。

@echo off
setlocal enabledelayedexpansion
set mdebug=0
if %mdebug%==1 (set "trace=echo") else (set "trace=rem")
%trace%.调试
echo.鉴于排版和显示问题，输入的数字不要大于9
echo.
set /p input=你想要的螺旋方阵是几阶的？请输入任意数字: 
set /p input2=顺时针还是逆时针？顺时针输入1，逆时针输入2: 

for /l %%i in (1,1,%input%) do (
  for /l %%j in (1,1,%input%) do (
  set "数组[%%i][%%j]=0"
  )
)

set "方向=0"
set /a "数组长=input*input"
set "行=1"
set "列=1"
set "数组[%行%][%列%]=01"

for /l %%j in (2,1,%数组长% ) do (

if !方向!==0 (
if !列!==%input% (set "bool1=1")  else  (set "bool1=0")
set /a "t=!列!+1"
for /f "delims=" %%a in ("数组[!行!][!t!]") do (set "t2=!%%a!")
if !t! leq %input% (if !t2! neq 0 (set "bool2=1")  else  (set "bool2=0"))
set /a "bool3=bool1|bool2"
if !bool3!==1 (
set /a "方向=(!方向!+1)%%4"
))

if !方向!==1 (
if !行!==%input% (set "bool1=1")  else  (set "bool1=0")
set /a "t=!行!+1"
for /f "delims=" %%a in ("数组[!t!][!列!]") do (set "t2=!%%a!")
if !t! leq %input% (if !t2! neq 0 (set "bool2=1")  else  (set "bool2=0"))
set /a "bool3=bool1|bool2"
if !bool3!==1 (
set /a "方向=(!方向!+1)%%4"
))

if !方向!==2 (
if !列!==1 (set "bool1=1")  else  (set "bool1=0")
set /a "t=!列!-1"
for /f "delims=" %%a in ("数组[!行!][!t!]") do (set "t2=!%%a!")
if !t! geq 1 (if !t2! neq 0 (set "bool2=1")  else  (set "bool2=0"))
set /a "bool3=bool1|bool2"
if !bool3!==1 (
set /a "方向=(!方向!+1)%%4"
))

if !方向!==3 (
if !行!==1 (set "bool1=1")  else  (set "bool1=0")
set /a "t=!行!-1"
for /f "delims=" %%a in ("数组[!t!][!列!]") do (set "t2=!%%a!")
if !t! geq 1 (if !t2! neq 0 (set "bool2=1")  else  (set "bool2=0"))
set /a "bool3=bool1|bool2"
if !bool3!==1 (
set /a "方向=(!方向!+1)%%4"
))

if !方向!==0 ( set /a "列=!列!+1" ) 
if !方向!==1 ( set /a "行=!行!+1" )
if !方向!==2 ( set /a "列=!列!-1" )
if !方向!==3 ( set /a "行=!行!-1" )
set /a "t3=%%j+100"
set "数组[!行!][!列!]=!t3:~1!"
)

if %input2%==2 (
for /l %%i in (1,1,%input%) do (
  for /l %%j in (1,1,%input%) do (
  (set /p "=!数组[%%j][%%i]!  ") < nul
  )
echo.
)
) else (
for /l %%i in (1,1,%input%) do (
  for /l %%j in (1,1,%input%) do (
  (set /p "=!数组[%%i][%%j]!  ") < nul
  )
echo.
)
)

pause >nul
复制代码

关于为啥数组要 加[][]是因为当n>11时，变量会重复比如:
数组111你说，到底是11行1列还是1行11列？


下面是我的第二种方法：

@echo off
setlocal enabledelayedexpansion
set mdebug=0
if %mdebug%==1 (set "trace=echo") else (set "trace=rem")
%trace%.调试
echo.鉴于排版和显示问题，输入的数字不要大于9
set /p n=你想要的螺旋方阵是几阶的？请输入任意数字: 
set /p input=顺时针还是逆时针？顺时针输入1，逆时针输入2: 
echo.
set /a "nn=n*n"
set "i=1"&set "j=1"
set /a "圈=(n+1)/2"

set "阶=!n!"
set "首位数值=1"
set "首位=1"
for /l %%a in (1,1,%圈%) do (
call :构造边缘 !首位数值! !阶! !首位!
set /a "首位数值+=(阶-1)*4"
set /a "阶-=2"
set /a "首位+=1"
)



if %input%==2 (
for /l %%i in (1,1,%n%) do (
  for /l %%j in (1,1,%n%) do (
  (set /p "=!数组[%%j][%%i]!  ") < nul
  )
echo.
)
) else (
for /l %%i in (1,1,%n%) do (
  for /l %%j in (1,1,%n%) do (
  (set /p "=!数组[%%i][%%j]!  ") < nul
  )
echo.
)
)
%trace%.调试
pause >nul
goto :eof

:构造边缘
if %2==1 (
set /a "t=%1+100"
set "数组[%3][%3]=!t:~1!"
) else (
set /a "圈长=(%2-1)*4"
set "i=%3"&set "j=%3"
set /a 末尾数值=%1+!圈长!-1
set "头=%3"
set /a "尾=头+%2-1"
for /l %%b in (%1,1,!末尾数值!) do (
set /a "t=%%b+100"
set "数组[!i!][!j!]=!t:~1!"
set "key1=1"
if !key1!==1 (if !i!==!头! (if !j! lss !尾! (
set /a j+=1
set "key1=0"
)))
if !key1!==1 (if !j!==!尾! (if !i! lss !尾! (
set /a i+=1
set "key1=0"
)))
if !key1!==1 (if !i!==!尾! (if !j! gtr !头! (
set /a j-=1
set "key1=0"
)))
if !key1!==1 (if !j!==!头! (if !i! gtr !头! (
set /a i-=1
set "key1=0"
)))
)
)
复制代码