Выведите числа, которые встречаются в списке несколько раз

a = [1, 2, 3, 2, 1, 5, 6, 5, 5, 5]
seen = set(x for x in a)
print(*seen)

a = [1, 2, 3, 2, 1, 5, 6, 5, 5, 5]
result = set(filter(lambda x: a.count(x) > 1,a))
print(*result)

import collections
a = [1, 2, 3, 2, 1, 5, 6, 5, 5, 5]
print([item for item, count in collections.Counter(a).items() if count > 1])

for x in a:
    if x not in seen:
        seen[x] = 1
    else:
        if seen[x] == 1:
            dupes.append(x)
        seen[x] += 1

a = [1, 2, 3, 2, 1, 5, 6, 5, 5, 5]
result = [e for e in set(a) if a.count(e)==1]

a = [1, 2, 3, 2, 1, 5, 6, 5, 5, 5]
result = set(e for i, e in enumerate(a) if e in a[i + 1:])
print(result)

In [1]: a = list(range(1_000))
 
In [2]: %timeit [e for e in set(a) if a.count(e)==1]
21.3 ms ± 422 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
 
In [3]: from collections import Counter
 
In [4]: %timeit [key for key, count in Counter(a).items() if count > 1]
220 µs ± 34.9 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

a = map(int, input().split())
used = set()
result = set()
for x in a:
    if x in used and x not in result:
        result.add(x)
    used.add(x)
    
print(*result)