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I want to write a combination of three-digit numbers (such as 001, 002...999), excluding combinations where two digits are equal (such as 001, 112, 313, 449, etc.) and combinations where all three digits are equal (such as 000, 111...999).

Note: For 012, 021, 120, 102, 210, 201, only one of them is taken, and the rest are similar.

Last edited by jmz573515 on 2007-1-26 at 06:41 PM ]

```vb
dim msg
combine "0123456789",3,""
msgbox msg
Sub Combine(sar,num,str)
If num=0 Then
msg=msg&str&";"
Else
For i=1 To Len(sar)
Call Combine(Right(sar,Len(sar)-i),num-1,str&Mid(sar,i,1))
Next
End If
End Sub
```

Okay, learned it.

(Currently restricted from adding points, will make up later.)

@echo off

set str=0123456789

cd.>result.txt

setlocal enabledelayedexpansion

for /l %%i in (0,1,7) do (

    set /a num1=%%i+1

    for /l %%j in (!num1!,1,8) do (

        set /a num2=%%j+1

        for /l %%k in (!num2!,1,9) do (

            set /a count+=1

            echo !str:~%%i,1!!str:~%%j,1!!str:~%%k,1!>>result.txt

        )

    )

)

echo   各个数位上的数字互不相同，且组成该数的数字与其他数中的数字不完全相同的三位数有 %count% 个>>result.txt 

start result.txt

I am using the idea of combinations with a BAT version:

@echo off

set str=0123456789

cd.>result.txt

setlocal enabledelayedexpansion

for /l %%i in (0,1,7) do (

    set /a num1=%%i+1

    for /l %%j in (!num1!,1,8) do (

        set /a num2=%%j+1

        for /l %%k in (!num2!,1,9) do (

            set /a count+=1

            echo !str:~%%i,1!!str:~%%j,1!!str:~%%k,1!>>result.txt

        )

    )

)

echo The three-digit numbers where the digits at each position are all different and the digits forming the number are not completely the same as those in other numbers have %count% in total>>result.txt 

start result.txt

Following ccwan's suggestion, the statistical result is added.

Last edited by namejm on 2007-1-26 at 11:08 PM ]

Banban's code is always so wonderful, appreciate it~~

The second floor is written in C, relatively simple! I like the code given by the fourth floor, thank you two!