中国DOS联盟

It will be calculated out, but not in one place. The form it gives is like this:
1×54=54
2×27=54
.......
2×27=54
3×18=54
……
3×18=54
6×9=54

Isn't that going to be a mess?

#25 ？
Number of chickens and dogs =
There are seventy-nine chickens and dogs, with two hundred feet on the ground. Think and calculate, how many chickens? How many dogs?

@echo off
set /a 鸡狗=79
set /a 鸡腿狗腿=200
set /a 最大狗=200/4
setlocal enabledelayedexpansion
for /l %%i in (1 1 %最大狗%) do (
set /a 狗腿=%%i*4
set /a 鸡腿=%鸡腿狗腿%-!狗腿!
set /a 鸡=!鸡腿!/2
set /a num=%%i+!鸡!
call :test !num! %%i !鸡! %鸡狗%
)

pause>nul

:test
if %1 EQU %4 echo 可能的组合为：狗=%2 鸡=%3
goto :eof

A good post should be topped up~:)


The way to top up a long-sunken good post without posting too much junk posts is: delete your original reply in the post (-1 point), and then reply again~:)


Last edited by redtek on 2007-1-28 at 07:14 PM ]

Posting up also has such knowledge, learned.

All are experts, drifting by```

So the original maximum number is 2^31 - 1. No wonder 2^31 overflows!

It would be nice if we could filter the multiples of 3, 4, 5 and 5, 12, 13.

The code on floor 84 is pretty good, and the running efficiency is very high.

@echo off
rem x+y=n m*x+y/m=n
set n=100
set m=4
set x=100/(1+m)
set y=x*m
echo Number of senior monks is %x%
echo Number of senior monks is %y%
pause>nul

There are several errors. The fifth line should change "100" to "n". The eighth line should change "大" to "small". This correction is hereby made. Please forgive me a lot.

### Step 1: Analyze the conditions
We can find that if we add 1 person to the number of people participating in the sports meeting, then the number of people can be exactly divided by 2, 3, 4, 5, and 6.

### Step 2: Find the least common multiple of 2, 3, 4, 5, and 6
- Prime factorize each number:
- \(2 = 2\)
- \(3 = 3\)
- \(4 = 2\times2\)
- \(5 = 5\)
- \(6 = 2\times3\)
- The least common multiple is the product of the highest powers of all prime factors involved. So the least common multiple of 2, 3, 4, 5, and 6 is \(2\times2\times3\times5= 60\)

### Step 3: Find the number of people participating in the sports meeting
Since adding 1 person makes it divisible by 60, then the number of people participating in the sports meeting is \(60 - 1 = 59\)

So the school has at least 59 people participating in this sports meeting.
The translated text is:### Step 1: Analyze the conditions
We can find that if we add 1 person to the number of people participating in the sports meeting, then the number of people can be exactly divided by 2, 3, 4, 5, and 6.

### Step 2: Find the least common multiple of 2, 3, 4, 5, and 6
- Prime factorize each number:
- \(2 = 2\)
- \(3 = 3\)
- \(4 = 2\times2\)
- \(5 = 5\)
- \(6 = 2\times3\)
- The least common multiple is the product of the highest powers of all prime factors involved. So the least common multiple of 2, 3, 4, 5, and 6 is \(2\times2\times3\times5 = 60\)

### Step 3: Find the number of people participating in the sports meeting
Since adding 1 person makes it divisible by 60, then the number of people participating in the sports meeting is \(60 - 1 = 59\)

So the school has at least 59 people participating in this sports meeting.

@echo off

setlocal enabledelayedexpansion

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

   title 正在检测 %%i

   set/a temp=%%i+1

   set/a flag=0

   for %%a in (2 3 4 5 6) do (

       set /a tmp=!temp! %% %%a

       if !tmp! EQU 0 set/a flag+=1)

   if !flag! GEQ 5 echo 该数字为：%%i & goto :exit)

:exit

pause>nul

Through mathematical knowledge, it is known that when this number is increased by 1, it is a multiple of 2, 3, 4, 5, 6.
Thus, the following code is available:

@echo off

setlocal enabledelayedexpansion

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

   title Detecting %%i

   set/a temp=%%i+1

   set/a flag=0

   for %%a in (2 3 4 5 6) do (

       set /a tmp=!temp! %% %%a

       if !tmp! EQU 0 set/a flag+=1)

   if !flag! GEQ 5 echo The number is: %%i & goto :exit)

:exit

pause>nul