中国DOS联盟

Not bad. Have research on algorithms.

Wow..In-depth...Learninging!!!

To promote algorithm research, heh heh.

There is a five-digit odd number. Replace all 2s with 5s and all 5s with 2s in this five-digit odd number, while keeping other digits unchanged, to get a new five-digit number. If half of the new five-digit number is still 1 more than the original five-digit number, what is the original five-digit number?

Haha... Interesting~ :)
Can start a batch processing and data structure algorithm special section~ :)

The sum of five consecutive natural numbers can be divided by 2, 3, 4, 5, and 6 respectively. Find the smallest set of such numbers that meets the condition.

Not bad~~After seeing so many treatments~~I think the several treatments upstairs are quite beautiful~~worthy of everyone's joint research~~Thank you all~~~

It is suggested that zouzhxi concentrate all issues in the top in the format of serial number + content listed. Once solved, make a mark or instruction behind this serial number, so that everyone can see which interesting topics are in this post.

@echo off & echo 正在检测数据...

setlocal enabledelayedexpansion

for /l %%a in (0 1 9) do (

   for /l %%b in (0 1 9) do (

     for /l %%c in (0 1 9) do (

       set/a Fnum=2%%a%%b%%c5

       set Bnum=!Fnum:5=A!

       set Bnum=!Bnum:2=5!

       set/a Bnum=!Bnum:A=2!

       set/a Fnum_=2*!Fnum!+2

       if !Bnum! EQU !Fnum_! echo !Fnum!

       )))

echo 检测完毕!

pause>nul

Finally back again!

Reply to post 18, the answer is: 29995.

The test code given:

@echo off & echo Checking data...

setlocal enabledelayedexpansion

for /l %%a in (0 1 9) do (

   for /l %%b in (0 1 9) do (

     for /l %%c in (0 1 9) do (

       set/a Fnum=2%%a%%b%%c5

       set Bnum=!Fnum:5=A!

       set Bnum=!Bnum:2=5!

       set/a Bnum=!Bnum:A=2!

       set/a Fnum_=2*!Fnum!+2

       if !Bnum! EQU !Fnum_! echo !Fnum!

       )))

echo Checking completed!

pause>nul

Explanation: This program was tested and passed under XP pro sp2.

To solve this problem, first we need to clarify that the number must end with 5, because only in this way, after conversion, it will be an even number and can be divided by 2; and the first digit must be 2......

Last edited by youxi01 on 2006-11-16 at 11:38 PM ]

@echo off

echo 正在检测数据....

setlocal enabledelayedexpansion

for /l %%i in (1 1 10000) do (

   set /a flag=0

   set /a num=5*%%i+10

   set /a num1=!num!%%4

   set /a num2=!num!%%5

   set /a num3=!num!%%6

   for %%a in (!num1! !num2! !num3!) do (

      if %%a NEQ 0 set /a flag=1 

    )

   if !flag! EQU 0 set /a num=%%i & goto :exit

)

:exit

for /l %%i in (0 1 4) do (

   set /a num%%i=%num%+%%i

)

echo 要求的连续的5个自然数为：%num0% %num1% %num2% %num3% %num4%

pause>nul

Answer to post 20: The answer is 10 11 12 13 14

@echo off

echo Detecting data....

setlocal enabledelayedexpansion

for /l %%i in (1 1 10000) do (

   set /a flag=0

   set /a num=5*%%i+10

   set /a num1=!num!%%4

   set /a num2=!num!%%5

   set /a num3=!num!%%6

   for %%a in (!num1! !num2! !num3!) do (

      if %%a NEQ 0 set /a flag=1 

    )

   if !flag! EQU 0 set /a num=%%i & goto :exit

)

:exit

for /l %%i in (0 1 4) do (

   set /a num%%i=%num%+%%i

)

echo The required consecutive 5 natural numbers are: %num0% %num1% %num2% %num3% %num4%

pause>nul

Tested and passed under XP Pro SP2.

Last edited by youxi01 on 2006-11-16 at 11:38 PM ]

At the request of namejm:

#Serial number! Solved? Unsolved

#01!
Find narcissistic numbers?

#02!
There are four numbers. When any three numbers are added together, the resulting sums are 84, 88, 99, 110 respectively. Find these four numbers?

#03!
Miss Zhao's age has the following characteristics:
1. Its cube is a four-digit number, and its fourth power is a six-digit number;
2. These four-digit number and six-digit number exactly consist of the ten digits from 0 to 9.
Ask, what number should this be?

#04!
A total of 4889 digits were used to number the pages of a dictionary. How many pages does this dictionary have? Answer: 1499

#05!
Acong said that he saw a group of camels in the northwest this time. There are 23 humps and 60 feet. How many single-humped and double-humped camels are there respectively?

#06!
There is a five-digit odd number. Replace all 2s with 5s and all 5s with 2s in this five-digit odd number, and keep other numbers unchanged to get a new five-digit number. If half of the new five-digit number is still 1 greater than the original five-digit number, what is the original five-digit number?

#07!
The sum of five consecutive natural numbers can be divisible by 2, 3, 4, 5, 6 respectively. Find the smallest set of numbers that meets this condition.

#08!
I am a three-digit number. There is one digit that is "3", another digit is "1", and the other digit is unknown. If "3" is changed to "4" and "1" is changed to "3", then the original me will be 9 less than half of the assumed me. Do you know what I was originally?

#09?
Farmer Jones said to his wife: "Hey, Maria, if I sell 75 chicks according to my way, then our chicken feed can last for 20 days. However, if I follow your suggestion and buy 100 more chicks, then the chicken feed will only last for 15 days."
"Ah, dear," she replied, "then how many chicks do we have now?"
The problem is here. How many chicks do they have exactly?

#10!
Among all five-digit numbers, how many have exactly two 3s?

#11!
Divide 17 into the sum of several natural numbers. What is the maximum product of these natural numbers?

#12?
Multiply natural numbers 2, 3... together. The last 6 digits of their product are exactly all 0s. What is the minimum possible value of the last natural number?

#13!
The sum of the dividend, divisor and quotient is 181, and the quotient is 12. Find the dividend.

#14?
There are six boxes of goods in the store, weighing 15, 16, 18, 19, 20, 31 kilograms respectively. Two customers bought five of them. It is known that the weight of the goods bought by one customer is twice that of the other customer. Then, what is the weight of the remaining box of goods in the store?

#15?
A number has a remainder of 2 when divided by 3 and a remainder of 1 when divided by 5. What is the remainder of this number when divided by 15?

#16?
1. p is a prime number, and p×p + 1 is also a prime number. Find 2006×p.
2. What is the remainder when the product of 2006 2s is divided by 7?

Last edited by zouzhxi on 2006-11-15 at 09:39 PM ]

@echo off

setlocal enabledelayedexpansion

for /l %%i in (10001,2,99999) do (

    set num=%%i

    set num=!num:5=a!

    if not "!num!"=="%%i" (

        set num_tmp1=!num!

        set num=!num:2=5!

        if not "!num!"=="!num_tmp1!" (

            set num=!num:a=2!

            set /a num_tmp2=2*%%i+2

            if !num! equ !num_tmp2! echo %%i

        )

    )

)

pause

@echo off

setlocal enabledelayedexpansion

for /l %%i in (10005,5,99995) do (

    set num=%%i

    set num=!num:5=a!

    if not "!num!"=="%%i" (

        set num_tmp1=!num!

        set num=!num:2=5!

        if not "!num!"=="!num_tmp1!" (

            set num=!num:a=2!

            set /a num_tmp2=2*%%i+2

            if !num! equ !num_tmp2! echo %%i

        )

    )

)

pause

@echo off

setlocal enabledelayedexpansion

for /l %%i in (20005,5,29995) do (

    set num=%%i

    set num=!num:5=a!

    set num=!num:2=5!

    set num=!num:a=2!

    set /a num_tmp=2*%%i+2

    if !num! equ !num_tmp! echo %%i

)

pause

Originally posted by youxi01 at 2006-11-14 18:24:
To solve this problem, first we need to clarify that the number must end with 5, because only in this way can it be an even number after conversion and be divisible by 2; and the first digit must be 2...

　　The idea of 23F has flaws, because what we need to do is to simulate the problem statement with code. If we can directly deduce certain conditions, we can make full use of these conditions; if the conditions need to be deduced through more than two steps of reasoning, it is better not to use them and should be left to the code to handle; otherwise, since the conditions deduced through two steps of reasoning can be used, then what about three steps? Four steps? N steps? The last step is just included in the Nth step? In that case, we can completely write out the conclusion that needs to be deduced through N steps and thus completely abandon the code - although your result is correct in the end, it is suggested not to do this.

　　Based on 23F, here is a code that solves the problem of 18F and tries to simulate the problem statement (very slow):

@echo off

setlocal enabledelayedexpansion

for /l %%i in (10001,2,99999) do (

    set num=%%i

    set num=!num:5=a!

    if not "!num!"=="%%i" (

        set num_tmp1=!num!

        set num=!num:2=5!

        if not "!num!"=="!num_tmp1!" (

            set num=!num:a=2!

            set /a num_tmp2=2*%%i+2

            if !num! equ !num_tmp2! echo %%i

        )

    )

)

pause

　　Then here is a faster code:

@echo off

setlocal enabledelayedexpansion

for /l %%i in (10005,5,99995) do (

    set num=%%i

    set num=!num:5=a!

    if not "!num!"=="%%i" (

        set num_tmp1=!num!

        set num=!num:2=5!

        if not "!num!"=="!num_tmp1!" (

            set num=!num:a=2!

            set /a num_tmp2=2*%%i+2

            if !num! equ !num_tmp2! echo %%i

        )

    )

)

pause

　　The code of 23F can be simplified as:

@echo off

setlocal enabledelayedexpansion

for /l %%i in (20005,5,29995) do (

    set num=%%i

    set num=!num:5=a!

    set num=!num:2=5!

    set num=!num:a=2!

    set /a num_tmp=2*%%i+2

    if !num! equ !num_tmp! echo %%i

)

pause

Last edited by namejm on 2006-11-15 at 12:57 PM" ]

I can't agree with the statement of the "moderator" above: "Because what we need to do is to simulate the problem with code. If we can directly derive some conditions, we can make full use of these conditions; if conditions that need more than two steps of reasoning to be obtained".

I once wrote a problem of finding prime numbers within 100,000 with (pure) batch processing. According to the statement above, if we simulate the problem, the running efficiency of the written code is really not to be praised!

For some derivable conclusions, especially those that can be derived quickly, I think they can all be applied.

Hehe, the approach of not agreeing is correct because I didn't state my precondition: when we want to study certain algorithms, we should try to write code simulating the problem statement; if it's writing scripts, based on the principle of , we can make full use of all correctly derivable conditions - the reason for proposing the statement of trying to simulate here is because we are mainly discussing general algorithms here. Of course, it doesn't mean excluding highly technical code.

　　When playing number games with batch processing, you need to pay attention to the range of numbers, otherwise overflow will occur.

　　-2147483648——2147483647 （WINDOWS XP SP2 @ CMD Shell）

@echo off

rem 经推算，首位数字必须为1，（除非未知数为1，为1的情况在第一个for中也处理了）

setlocal enabledelayedexpansion

for /l %%i in (0 1 9) do (

   set /a Fnum=1%%i3

   set /a Bnum=3%%i4

   set /a num=!Bnum!/2-9

   if !Fnum! EQU !num! echo !Fnum!)



for /l %%i in (0 1 9) do (

   set /a Fnum=13%%i

   set /a Bnum=34%%i

   set /a num=!Bnum!/2-9

   if !Fnum! EQU !num! echo !Fnum!)

pause>nul

@echo off

rem 小鸡的只数至少有76只:



setlocal enabledelayedexpansion

for /l %%i in (76 1 10000) do (

   set /a Fnum=%%i*20-75*20

   set /a Bnum=%%i*15+100*15

  if !Fnum! EQU !Bnum! echo 小鸡的只数为：%%i & goto :exit)

:exit

pause>nul

@echo off

echo 正在检测数据........

setlocal enabledelayedexpansion

set /a flag=0

for /l %%i in (10000,1,99999) do (

   set num=%%i

rem 前面加个1是为了防止类似30820的特殊情况。

   set /a num=1!num:3=!

   if !num! lss 2000 if !num! gtr 200 set /a flag+=1)

echo %flag%

echo 检测完毕!

pause>nul

Not bad, there are many questions! Let's answer a few first!

### Question 8:
```
@echo off
rem After deduction, the first digit must be 1, (unless the unknown number is 1, and the case where it is 1 is also handled in the first for)
setlocal enabledelayedexpansion
for /l %%i in (0 1 9) do (
set /a Fnum=1%%i3
set /a Bnum=3%%i4
set /a num=!Bnum!/2-9
if !Fnum! EQU !num! echo !Fnum!)

for /l %%i in (0 1 9) do (
set /a Fnum=13%%i
set /a Bnum=34%%i
set /a num=!Bnum!/2-9
if !Fnum! EQU !num! echo !Fnum!)
pause>nul

```

### Question 9:
```
@echo off
rem The number of chicks is at least 76:

setlocal enabledelayedexpansion
for /l %%i in (76 1 10000) do (
set /a Fnum=%%i*20-75*20
set /a Bnum=%%i*15+100*15
if !Fnum! EQU !Bnum! echo The number of chicks is: %%i & goto :exit)
:exit
pause>nul

```

### Question 10: Here is a relatively clever method: (the efficiency may be relatively high, not tested!)
```
@echo off
echo Detecting data........
setlocal enabledelayedexpansion
set /a flag=0
for /l %%i in (10000,1,99999) do (
set num=%%i
rem Adding a 1 in front is to prevent special cases like 30820.
set /a num=1!num:3=!
if !num! lss 2000 if !num! gtr 200 set /a flag+=1)
echo %flag%
echo Detection completed!
pause>nul
```

Last edited by youxi01 on 2006-11-16 at 11:41 PM ]