1,如何使嵌套时的循环处理更快。从(a)到(zzzzz)处理1个字符需要30秒
2.如何使此代码能够设置范围,例如以2位开始到4位结束。
3.为什么此代码显示编码命令,为什么我在命令中放置字符@
Sample.bat
@echo off
setlocal EnableDelayedExpansion
set "loop2=IN (0,1,36) DO @("
set "n0="
set "n1=a"
set "n2=b"
set "n3=c"
set "n4=d"
set "n5=e"
set "n6=f"
set "n7=g"
set "n8=h"
set "n9=i"
set "n10=j"
set "n11=k"
set "n12=l"
set "n13=m"
set "n14=n"
set "n15=o"
set "n16=p"
set "n17=q"
set "n18=r"
set "n19=s"
set "n20=t"
set "n21=u"
set "n22=v"
set "n23=w"
set "n24=x"
set "n25=y"
set "n26=z"
set "n27=1"
set "n28=2"
set "n29=3"
set "n30=4"
set "n31=5"
set "n32=6"
set "n33=7"
set "n34=8"
set "n35=9"
set "n36=0"
@FOR /L %%a %loop2%
@call :prc %%a
@FOR /L %%b %loop2%
@call :prc %%a %%b
@FOR /L %%c %loop2%
@call :prc %%a %%b %%c
@FOR /L %%d %loop2%
@call :prc %%a %%b %%c %%d
@FOR /L %%e %loop2%
@call :prc %%a %%b %%c %%d %%e
)
)
)
)
)
)
@pause
:prc
::@cls
@title %1 %2 %3 = !n%1!!n%2!!n%3!
@set "data=!n%1!!n%2!!n%3!!n%4!!n%5!"
@if not defined data goto end
@echo.
@echo %data% >> data.txt
@echo.
@:end
@goto :eof
`
嵌套时如何使循环处理更快?
删除所有不必要的命令。
如何使此代码能够设置范围?
采用不同的递归方法构建字符串,
就像处理diff一样。数字基数。
BTW 36的5的幂是60.466.176,通常您以
数字0开头,后跟九个字母,如十六进制0-9abcdef
为什么此代码显示编码命令?
不会在这里发生Echo off
禁用文件输出,仅写入控制台并以第一个字母开头。
@echo off & setlocal EnableDelayedExpansion
set "n= abcdefghijklmnopqrstuvwxyz1234567890"
for /l %%n in (1,1,36) do set "n%%n=!n:~%%n,1!"
set "n="
set "loop2=in (1,1,36) DO ("
FOR /L %%a %loop2%
call :prc %%a
FOR /L %%b %loop2%
call :prc %%a %%b
FOR /L %%c %loop2%
call :prc %%a %%b %%c
FOR /L %%d %loop2%
call :prc %%a %%b %%c %%d
FOR /L %%e %loop2%
call :prc %%a %%b %%c %%d %%e
)
Pause
)
)
)
)
pause
Goto :Eof
:prc
set "data=!n%1!!n%2!!n%3!!n%4!!n%5!"
echo %data% &Rem>> data.txt
goto :eof
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