Submission #10261933

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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
from collections import Counter
if Counter(D).most_common(1)[0][1] >= 3:
    print(0)
    exit()

#同じ数が2個以下なので、Nは26以下
#2**26パターンを全調査する
ans = 0
import heapq
for pattern in range(1<<N):
    q = []
    for d in D:
        if pattern & 1: d = 24 - d
        heapq.heappush(q, d)
        pattern>>=1
    pre = 0
    tmp = 24
    for _ in range(N):
        d = heapq.heappop(q)
        tmp = min(tmp, abs(pre-d))
        pre = d
    ans = max(ans, tmp)

print(ans)

Submission Info

Judge Result