China DOS Union

The problem is like this: there is a four-digit number. The first digit is the number of 0s in this number, the second digit is the number of 1s in this number, the third digit is the number of 2s in this number, and the fourth digit is the number of 3s in this number. Find this number.

- - ||| I actually didn't even understand what it means...

Olympiad math problems probably shouldn't be done with exhaustive search, otherwise the Olympiad math teacher would harshly criticize you :)

"The first digit is the number of 0s in this number", so the first digit definitely cannot be 0, otherwise it's a contradiction.

Not sure whether the units digit is the first digit, or the thousands digit is the first digit.

Reply #3 is correct, but it doesn't seem to have been calculated with P.
The first digit cannot be 0, otherwise it would be a three-digit number.
Also, that's if "the first digit" in the problem means counting from the left. If the first digit means counting from the right, then I don't know what the answer is. Still hope it can be done with P

If the first digit is counted from the right, then it's 0.003

The first digit means counting from the left. Answers: 1210 2020
If the first digit is counted from the right, then 0202 0121. If the first digit cannot be 0, otherwise it would be a three-digit number, then there is no solution.
Below is for counting the first digit from the left
@echo off
SETLOCAL ENABLEDELAYEDEXPANSION
for /l %%a in (0,1,3) do (
for /l %%b in (0,1,3) do (
for /l %%c in (0,1,3) do (
for /l %%d in (0,1,3) do (
set n=0
for /l %%e in (0,1,3) do (
set m%%e=0
if %%a==%%e set /a m%%e+=1
if %%b==%%e set /a m%%e+=1
if %%c==%%e set /a m%%e+=1
if %%d==%%e set /a m%%e+=1
)
if !m0!==%%a set /a n+=1
if !m1!==%%b set /a n+=1
if !m2!==%%c set /a n+=1
if !m3!==%%d set /a n+=1
if !n!==4 echo %%a%%b%%c%%d
))))
ENDLOCAL
pause

Shouldn't the outermost for start from 0?