Randomly display 5 different numbers - China DOS Union Forum

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

@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.

Floor1　26933062	Posted 2007-01-04 16:12
银牌会员　Posts 879　Credits 2,268
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 ]

Floor2　a9319751	Posted 2007-01-04 20:42
中级用户　Posts 170　Credits 439
5 digits? 5-digit? 1-100 has 5 digits

Floor3　zh159	Posted 2007-01-04 22:17
金牌会员　Posts 1,467　Credits 3,687
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 ]

Floor4　youxi01	Posted 2007-01-04 23:45
高级用户　Posts 247　Credits 846　From 湖南==》广东
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

Floor5　namejm	Posted 2007-01-05 01:52
荣誉版主　Posts 1,737　Credits 5,226　From 成都
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.

Floor6　namejm	Posted 2007-01-05 02:16
荣誉版主　Posts 1,737　Credits 5,226　From 成都
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 ]

Floor7　tghksj	Posted 2007-01-05 03:05
社区乞丐　Posts 90　Credits -49
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!

Floor8　namejm	Posted 2007-01-05 03:33
荣誉版主　Posts 1,737　Credits 5,226　From 成都
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.

Floor9　tghksj	Posted 2007-01-05 04:30
社区乞丐　Posts 90　Credits -49
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!

Floor10　ccwan	Posted 2007-01-05 04:36
金牌会员　Posts 1,160　Credits 2,725　From 河北廊坊
Do the 5 numbers obtained include duplicates?

Floor11　namejm	Posted 2007-01-05 04:41
荣誉版主　Posts 1,737　Credits 5,226　From 成都
According to the building楼主's statement, there should be no duplicate numbers among the 5 numbers.

Floor12　ccwan	Posted 2007-01-05 04:43
金牌会员　Posts 1,160　Credits 2,725　From 河北廊坊
At present, this situation has been found on the 9th floor.

Floor13　tghksj	Posted 2007-01-05 06:18
社区乞丐　Posts 90　Credits -49
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 ]

Floor14　tghksj	Posted 2007-01-05 06:44
社区乞丐　Posts 90　Credits -49
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 ]

Floor15　namejm	Posted 2007-01-05 06:58
荣誉版主　Posts 1,737　Credits 5,226　From 成都
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.