Реверс числа

def rev(n):
    r=0
    while(n>0):
        r=r*10+n%10
        n=n//10
    return r
 
n=int(input())
r=rev(n)
print(n)

print(int(str(input())[::-1]))

def rzv(x):
    return int("".join(reversed(str(x))))

def reverse(n):
    result = 0
    while n > 0:
        n, r = divmod(n, 10)
        result = 10 * result + r
    return result
    
n = int(input())
result = reverse(n)
print(result)

In [1]: n = 123
 
In [2]: def rev(n):
   ...:     r=0
   ...:     while(n>0):
   ...:         r=r*10+n%10
   ...:         n=n//10
   ...:     return r
   ...:
 
In [3]: %timeit rev(123)
1.2 µs ± 102 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
 
In [4]: %timeit int(str(n)[::-1])
880 ns ± 5.22 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
 
In [5]: def rzv(x):
   ...:     return int("".join(reversed(str(x))))
   ...:
 
In [6]: %timeit rzv(n)
1.54 µs ± 18.5 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
 
In [7]: def reverse(n):
   ...:     result = 0
   ...:     while n > 0:
   ...:         n, r = divmod(n, 10)
   ...:         result = 10 * result + r
   ...:     return result
   ...:
 
In [8]: %timeit reverse(n)
1.35 µs ± 158 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

numbers = map(int, input().split())
reversed_numbers = [int(str(num)[::-1]) for num in numbers]
print(*reversed_numbers[::-1])