Submission #10262333

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Source Code Expand

N=int(input())
D=list(map(int, input().split()))

#同じ数が3個以上あるとsは0になる
#0か12が2個以上ある場合も0になる
from collections import Counter
C = Counter(D)
if C.most_common(1)[0][1] >= 3 or\
    C[0] > 1 or C[12] > 1:
    print(0)
    exit()

#同じ数が2個以下、0と12は1個以下なので、Nは24以下
#2**24パターンを全調査する

def diff(a, b):
    d = abs(a-b)
    return min(d, 24-d)

ans = 0
import heapq
for pattern in range(1<<N):
    q = []
    for d in D:
        if pattern & 1:
            if d == 0 or d == 12: break
            d = 24 - d
        heapq.heappush(q, d)
        pattern>>=1
    if len(q) != N: continue
    pre = 0
    tmp = 24
    for _ in range(N):
        d = heapq.heappop(q)
        tmp = min(tmp, diff(pre, d))
        pre = d
    ans = max(ans, tmp)

print(ans)

Submission Info

Judge Result