#976019 (Python3) No.2732 Similar Permutations

• 問題: No.2732 Similar Permutations
• ユーザー: titia
• 提出日時: 2024-04-21 01:52:50
• 言語: Python3 (3.13.1 + numpy 2.2.1 + scipy 1.14.1)
• 結果: WA
• 実行時間: -
• コード長: 2,214 bytes
• コンパイル時間: 91 ms
• コンパイル使用メモリ: 12,928 KB
• 実行使用メモリ: 46,732 KB
• 最終ジャッジ日時: 2024-10-13 00:20:40
• 合計ジャッジ時間: 40,995 ms
• ジャッジサーバーID （参考情報）: judge3 / judge2

ファイルパターン	結果
sample	AC * 2
other	AC * 73 WA * 28

ソースコード

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import sys
input = sys.stdin.readline
N=int(input())
A=list(map(int,input().split()))
if N==2:
    x=(A[0]+1)^(A[1]+2)
    y=(A[0]+2)^(A[1]+1)
    if x==y:
        print(1,2)
        print(2,1)
    else:
        print(-1)
    exit()
X=[[] for i in range(4)]
for i in range(N):
    X[A[i]%4].append(i)
for i in range(4):
    if i>=2 and len(X[i])>=2:
        k=X[i][0]
        l=X[i][1]
        ANS1=[0]*N
        ANS2=[0]*N
        ANS1[k]=2
        ANS1[l]=3
        ANS2[k]=3
        ANS2[l]=2
        LL=[2,3]
        now=1
        for i in range(N):
            if ANS1[i]==0:
                ANS1[i]=now
                now+=1
            while now in LL:
                now+=1
        now=1
        for i in range(N):
            if ANS2[i]==0:
                ANS2[i]=now
                now+=1
            while now in LL:
                now+=1
        print(*ANS1)
        print(*ANS2)
        exit()
    if i<=1 and len(X[i])>=2:
        k=X[i][0]
        l=X[i][1]
        ANS1=[0]*N
        ANS2=[0]*N
        ANS1[k]=2
        ANS1[l]=1
        ANS2[k]=1
        ANS2[l]=2
        LL=[1,2]
        now=1
        for i in range(N):
            if ANS1[i]==0:
                ANS1[i]=now
                now+=1
            while now in LL:
                now+=1
        now=1
        for i in range(N):
            if ANS2[i]==0:
                ANS2[i]=now
                now+=1
            while now in LL:
                now+=1
        print(*ANS1)
        print(*ANS2)
        exit()
from itertools import permutations
from collections import Counter
if N<=5:
    L=list(permutations(range(N)))
    ANS=[0]*len(L)
    for i in range(len(L)):
        xor=0
        for j in range(N):
            xor^=(A[j]+L[i][j]+1)
        ANS[i]=xor
    C=Counter(ANS)
    ans=-1
    for c in C:
        if C[c]>=2:
            ans=c
            break
    if ans==-1:
        print(ans)
        exit()
    IND=[]
    for i in range(len(L)):
        if ANS[i]==ans:
            IND.append(i)
    AA=[0]*N
    for i in range(N):
        AA[i]=L[IND[0]][i]+1
    print(*AA)
    AA=[0]*N
    for i in range(N):
        AA[i]=L[IND[1]][i]+1
    print(*AA)
    
    
 
 
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