China DOS Union

Can you display 5-digit numbers randomly from 1 to 100 at once without repetition?
I've thought about it for a long time and searched most of the posts in the forum, but it seems there are no posts that display 5-digit numbers at once, and the key is that there should be no repetition.


Sorry, I'm sorry everyone. It seems my expression ability is really problematic.
My purpose is to display random numbers from 1 to 100, display 5 at a time, and the 5 numbers displayed should not be repeated. Can everyone understand this?


First of all, thank you all for your wonderful answers. Because my batch processing level is really too low, I can't understand some advanced answers well, so please forgive me if there are any wrong evaluations.....................
1. The answer on the 6th floor basically meets the requirements, but there is an occasional dead loop phenomenon when used multiple times. I don't know the reason.
2. The one on the 13th floor has no repetition phenomenon, but the first digit is always the same every time it runs, and I always feel a little inappropriate.
3. The one on the 26th floor can also be used, but it feels too rigid. (I'm sorry, maybe the wording is inappropriate, but I can't think of a more euphemistic word, please forgive me!) For example, if I want to display 6, 7, 8, 9... digits, it's too troublesome. In fact, my original intention is to first require entering the maximum value of the random number, and then entering the number of digits to display.
4. Personally, I feel that (in terms of thinking) the one on the 32nd floor should be the most ideal, but unfortunately I can't understand it, and the brother on the 32nd floor didn't write the code as required. (Displayed 100 at once).
5. I think the code on the 33rd floor should be the best. Judging from the code, there should be absolutely no repetition, and it is very flexible. I just don't know if there will be the problems they said about low efficiency and dead loops. Currently, the running condition is good.
size]

Sorry, moderator. After repeated testing, your method on the 6th floor is okay, but there will be a dead loop. Maybe I tested too many methods and got the wrong object. I'm sorry. The method modified on the 42nd floor I think is close to perfect. There are so many good methods, batch processing is really a challenge to people's thinking
.

[ Last edited by 26933062 on 2007-1-6 at 10:07 PM ]

5 digits? 5-digit?
1-100 has 5 digits

5 randomly selected numbers from 1 to 99

1:

@echo off

:loop

cls

set random1=%random:~-2%



:loop2

set random2=%random:~-2%

if "%random2%" == "%random1%" goto loop2



:loop3

set random3=%random:~-2%

if "%random3%" == "%random1%" goto loop3

if "%random3%" == "%random2%" goto loop3



:loop4

set random4=%random:~-2%

if "%random4%" == "%random1%" goto loop4

if "%random4%" == "%random2%" goto loop4

if "%random4%" == "%random3%" goto loop4



:loop5

set random5=%random:~-2%

if "%random5%" == "%random1%" goto loop5

if "%random5%" == "%random2%" goto loop5

if "%random5%" == "%random3%" goto loop5

if "%random5%" == "%random4%" goto loop5



echo %random1% %random2% %random3% %random4% %random5%

pause

goto loop

2:

@echo off

:Start

cls

call :loop 1

call :loop 2

call :loop 3

call :loop 4

call :loop 5



echo %random1% %random2% %random3% %random4% %random5%

pause

set Tmp_=

goto :Start



:loop

set randomloop=%random:~-2%

for %%n in (%Tmp_%) do if "%randomloop%" == "%%n" goto loop

set random%1=%randomloop%

call set Tmp_=%%Tmp_%% %randomloop%

goto :eof

[ Last edited by zh159 on 2007-1-4 at 10:09 AM ]

Come to join in the "fun"

@echo off

setlocal enabledelayedexpansion

set /a a=1



:test

    set /a _num!a!=%random:~-2%

    if not defined !_num%a%!  set /a a+=1

    if !a! LSS 6 goto :test



echo %_num1% %_num2% %_num3% %_num4% %_num5% 



pause>nul

The expression of the owner is problematic. After reading for a long time, I still don't know whether it is to generate 1-99 different random 5-digit numbers or to randomly display 5 different numbers in 1-99. Please make the owner's expression clear.

After looking at the code on floor 4, it is inferred that the original intention of youxi01 was to display 5 different random numbers between 1 and 99 (excluding 99), but there is a problem with the code: %a% is a number from 1 to 5, and using if not defined to judge whether the variable num%a% has been defined, the execution result is always not defined.

Modify the code according to the idea of floor 4 as follows:

@echo off

set count=0



:loop

set /a num=%random%%%99+1

if not defined num%num% (

    set num%num%=%num%

    set /a count+=1

    call echo %%num%%

)

if %count% lss 5 goto loop



pause>nul

——————————————————————————————————————
If this code is called in a loop, it will fall into an infinite loop at the 20th time. For the specific situation, please refer to the description in floor 16. For the cause analysis, please refer to floor 41, and for the code to solve the problem, please refer to floor 42.

[ Last edited by namejm on 2007-1-6 at 05:03 PM ]

Doubt about my own comprehension ability and the LZ's expression ability

Is it 5 numbers or 5 digits? 5 numbers from 1 to 100 as follows:

@ECHO %dbg% OFF

setlocal ENABLEDELAYEDEXPANSION



FOR /L %%A IN (1,1,5) DO ECHO !RANDOM:~-2!

I have discovered that many people have overlooked such a detail: %random:~-2% might take the string "00". Although set /a num=%random:~-2% can convert the "00" string into a numerical value, turning it into 0, however, 0 is not what the original poster wanted.

Okay,
%random:~-2% doesn't work, then %random:~0,2%

Ensure there's no 0 in front

@ECHO %dbg% OFF

setlocal ENABLEDELAYEDEXPANSION



FOR /L %%A IN (1,1,5) DO ECHO !RANDOM:~0,2!

Do the 5 numbers obtained include duplicates?

According to the building楼主's statement, there should be no duplicate numbers among the 5 numbers.

At present, this situation has been found on the 9th floor.

OK!!!Nested recursion! This time it won't repeat, right??? Hehe

@ECHO %dbg% OFF

setlocal ENABLEDELAYEDEXPANSION

SET T=1



:PR

FOR /L %%A IN (%T%,1,5) DO (SET NO%%A=!RANDOM:~0,2! & SET /A H=%%A-1 & IF !H! GTR -1 (

FOR /L %%B IN (!H!,-1,1) DO (IF !NO%%A!==!NO%%B! (SET T=%%A & CALL :PR !T! & GOTO :EOF)))&ECHO !NO%%A!)

[ Last edited by tghksj on 2007-1-4 at 06:11 PM ]

Oh my god~~ I'm stupid, why didn't I think of if not defined
Dizzy dizzy dizzy dizzy dizzy dizzy dizzy dizzy dizzy dizzy dizzy dizzy dizzy dizzy dizzy

@ECHO %dbg% OFF

setlocal ENABLEDELAYEDEXPANSION

CLS

SET T=1



:PR

FOR /L %%A IN (%T%,1,5) DO (

         SET NO%%A=!RANDOM:~0,2!

         IF DEFINED !NO%%A! (SET T=!%%A! & GOTO :PR)

         ECHO !NO%%A!

                            )

Hey?????????? Doesn't work????????????? There are still duplicates??!
Experts explain it...

Only the 13th floor is available for the time being....

[ Last edited by tghksj on 2007-1-4 at 05:53 PM ]

After looking at the code on floor 14, I found that there is no statement to control whether random numbers are repeated. After testing several times, you will find that repeated values appear. It is recommended to refer to the code I posted on floor 6.