Question

批处理for/l命令

使用for /l 显示下面的数字:
11 21 22 31 32 33 41 42 43 44 51 52 53 54 55

不知道怎么排列出来的，请用批处理做出来，谢谢
注释就不用了，给出程序就OK

Answer 1

这个严格说来不算是一个编程题，只是一个找规律的。这样排开就容易开出来了：
11
21 22
31 32 33
41 42 43 44
51 52 53 54 55

Answer 2

@for /l %%n in (1,1,5) do @for /l %%a in (1,1,%%n) do @echo %%n%%a
::直接命令行下执行的话，则用一个%， 想必你知道的 ^_^

Answer 3

@echo off&mode concols=55lines=2&setlocal enabledelayedexpansion
set/a c=%random%%%10
color %c%a
set a=0
set setin=11 21 22 31 32 33 41 42 43 44 51 52 53 54 55
title %setin%
set setin1=55 54 53 52 51 44 43 42 41 33 32 31 22 21 11
set heads=
for /l %%a in (0)do (set/a a+=1, b=%a%
if !b!==0 (set step=11,-1,0&set step1=0,1,11)else set step=0,1,11&step1=11,-1,0
for /l %%1 in (!step1!)do (for /l %%2 in (1,1,%%1)do set/p= <nul
set/p=!setin!<nul&call:sleep&set/p=!heads!<nul&call:sleep&set/p=!heads!<nul&call:sleep)&echo.&echo.&echo.&cls
for /l %%1 in (!step!)do (for /l %%2 in (1,1,%%1)do set/p= <nul
set/p=!setin!<nul&call:sleep&set/p=!heads!<nul&call:sleep&set/p=!heads!<nul&call:sleep)&echo.&echo.&echo.&cls)
:sleep
::1000
ping/n 1 127.1>nul

Answer 4

@echo off&setlocal enabledelayedexpansion
for /l %%i in (1,1,5) do (
for /l %%j in (1,1,%%i) do set "hh=!hh! %%i%%j" )
echo %hh%
pause

Answer 5

@echo off&for /l %%a in (1 1 5) do for /l %%b in (1 1 %%a) do set/p ml=%%a%%b <nul
pause>nul

像一楼，但又不是