中国DOS联盟

As the title: Display with batch processing

1
121
12321
........................
12345678987654321
Prohibit using the echo command entirely.

@echo off

setlocal enabledelayedexpansion

set "space=         "



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

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

         set str1=!str1!%%a

         set str2=%%a!str2!

         set str=%space%!str1!!str2:~1!

      )

    echo !str:~%%i!

    set str1=

    set str2=

    set str=

 )

pause>nul

Try the following code?

@echo off

setlocal enabledelayedexpansion

set "space=         "



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

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

         set str1=!str1!%%a

         set str2=%%a!str2!

         set str=%space%!str1!!str2:~1!

      )

    echo !str:~%%i!

    set str1=

    set str2=

    set str=

 )

pause>nul

This is a very simple math problem. Mom bought a basket of apples. The eldest took 1/3, the second eldest took 1/3 of the remaining, the third one took 1/3 of the remaining. At this time, there are still 8 apples. How many apples are there in total?

As the title says, N >= 3, no need to display in table form. For example:

2 7 6

9 5 1

4 3 8

### Problem 1
Let the number of men be \(x\), the number of women be \(y\), and the number of children be \(z\). Then we have the following system of equations:
\





Multiply equation \((2)\) by \(2\) to get rid of the fraction: \(8x + 6y + z = 72 \quad (3)\)
Subtract equation \((1)\) from equation \((3)\): \((8x + 6y + z) - (x + y + z) = 72 - 36\)
\(7x + 5y = 36\)
Since \(x\) and \(y\) are positive integers, we can try values for \(x\):
If \(x = 1\), then \(7\times1 + 5y = 36\), \(5y = 29\), \(y\) is not an integer.
If \(x = 2\), then \(7\times2 + 5y = 36\), \(14 + 5y = 36\), \(5y = 22\), \(y\) is not an integer.
If \(x = 3\), then \(7\times3 + 5y = 36\), \(21 + 5y = 36\), \(5y = 15\), \(y = 3\)
Substitute \(x = 3\) and \(y = 3\) into equation \((1)\): \(3 + 3 + z = 36\), \(z = 30\)

So there are 3 men, 3 women, and 30 children.

### Problem 2
Let the number of bullocks be \(x\), the number of cows be \(y\), and the number of calves be \(z\). Then we have:
\





Simplify equation \((2)\) by dividing by 100: \(30x + 40y + 6z = 1000\), further divide by 2: \(15x + 20y + 3z = 500 \quad (3)\)
Multiply equation \((1)\) by 3: \(3x + 3y + 3z = 300 \quad (4)\)
Subtract equation \((4)\) from equation \((3)\): \((15x + 20y + 3z) - (3x + 3y + 3z) = 500 - 300\)
\(12x + 17y = 200\)
Solve for \(x\): \(x = \frac{200 - 17y}{12}\)
Since \(x\) and \(y\) are positive integers, we try values for \(y\):
If \(y = 4\), then \(x = \frac{200 - 17\times4}{12} = \frac{200 - 68}{12} = \frac{132}{12} = 11\)
Substitute \(x = 11\) and \(y = 4\) into equation \((1)\): \(11 + 4 + z = 100\), \(z = 85\)

So there are 11 bullocks, 4 cows, and 85 calves.

There is a group of people buying sheep together. If each person pays 5 taels, there is still a shortage of 45 taels. If each person pays 7 taels, there is still a shortage of 3 taels. Ask how many people are buying sheep and how much the sheep costs.

@echo off

setlocal enabledelayedexpansion



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

    title 正在检测 %%i ....

    set /a tmp=%%i*8/27

    if !tmp! EQU 8 echo %%i && goto :exit

 )

:exit

pause>nul

Originally posted by 6622186 at 2007-3-20 02:41 AM:
This is a very simple math problem. Mother bought a basket of apples. The eldest took 1/3, the second took 1/3 of the remaining, the third took 1/3 of the remaining, and there were 8 apples left. How many apples were there in total?

@echo off

setlocal enabledelayedexpansion



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

    title Checking %%i....

    set /a tmp=%%i*8/27

    if !tmp! EQU 8 echo %%i && goto :exit

 )

:exit

pause>nul

Actually, the above code can be simpler (according to mathematical principles), so simple that it's embarrassing (there's not much code to write).

This way, you can calculate numbers greater than 2^31. I thought it was impossible, but I don't understand the code. Seeing! gives me a headache.

112th floor

Did you first calculate it and then write it like this?

Your sentence "set/a tem=%%i*8/27&np if !temp! equ 8 echo %%i....." doesn't it mean that %%i is 27?

Then what are you calculating?

The complete solution is as follows:

Let the total be Y.

The first time, 1/3 of it is taken, leaving 2/3 * Y;

The second time, 1/3 of the remaining is taken again, leaving 2/3 * (2/3 * Y);

The third time, 1/3 of the remaining after the second time is taken, leaving only 2/3 * (2/3 * (2/3 * Y)).

hehe

This means: 2/3 * (2/3 * (2/3 * Y)) = 8

Then Y is 27.

Originally posted by scriptor at 2007-3-20 06:29 PM:
Post 112

Did you calculate first and then write it like this?

Your sentence "set/a tem=%%i*8/27&np if !temp! equ 8 echo %%i....." means that %%i is 27?

Then what are you calculating?

If we say, to completely simulate solving it with a program, it seems simpler. According to the meaning of the problem, this number must be a multiple of 27 (it can be continuously divided by 3 three times). So the code can be written as:

for /l %%i (27 27 100000) do (
....)

It seems simpler, and you can get 27 at once.

@echo off

setlocal enabledelayedexpansion

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

    for /l %%j in (9,-1,%%i) do set /p= <nul

    for /l %%k in (1,1,%%i) do set /p=%%k<nul

    set /a num=%%i-1

    for /l %%l in (!num!,-1,1) do set /p=%%l<nul

    echo.

)

pause

Originally posted by 6622186 at 2007-3-19 12:07:
As the title says: Display using batch processing

1
121
12321
........................
12345678987654321
Prohibit the use of all echo commands.

　　Before leaving, post a code:

@echo off

setlocal enabledelayedexpansion

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

    for /l %%j in (9,-1,%%i) do set /p= <nul

    for /l %%k in (1,1,%%i) do set /p=%%k<nul

    set /a num=%%i-1

    for /l %%l in (!num!,-1,1) do set /p=%%l<nul

    echo.

)

pause

@echo off
echo.
Chickens and dogs total 79, with 200 feet on the ground. Think and calculate, how many chickens? How many dogs?
echo.
setlocal enabledelayedexpansion
set /a x=79
set /a y=200
for /l %%i in (1,1,79) do (
for /l %%j in (1,1,50) do (

set /a a=2*%%i
set /a b=4*%%j
set /a sum1=%%i+%%j
set /a sum2=!a!+!b!
if !sum1! equ !x! if !sum2! equ !y! echo %%i %%j
)
)
pause

@echo off
echo.
One hundred monks and one hundred pastries. Each big monk gets four pastries, and four little monks share one pastry. How many big and little monks are there respectively?
echo.
setlocal enabledelayedexpansion
set /a x=100
for /l %%i in (1,1,25) do (
for /l %%j in (1,1,100) do (

set /a a=4*%%i
set /a b=%%j/4
set /a sum1=%%i+%%j
set /a sum2=!a!+!b!
if !sum1! equ !x! if !sum2! equ !x! echo "Big monk" %%i "Little monk" %%j
)
)
pause