結果
問題 |
No.1013 〇マス進む
|
ユーザー |
![]() |
提出日時 | 2023-03-31 10:52:19 |
言語 | Python3 (3.13.1 + numpy 2.2.1 + scipy 1.14.1) |
結果 |
TLE
|
実行時間 | - |
コード長 | 2,378 bytes |
コンパイル時間 | 329 ms |
コンパイル使用メモリ | 12,800 KB |
実行使用メモリ | 124,512 KB |
最終ジャッジ日時 | 2024-09-22 16:06:58 |
合計ジャッジ時間 | 29,698 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 33 TLE * 3 -- * 26 |
ソースコード
import sys readline=sys.stdin.readline class Path_Doubling: def __init__(self,N,permutation,lst=None,f=None,e=None): self.N=N self.permutation=permutation self.lst=lst self.f=f self.e=e def Build_Next(self,K=None): if K==None: self.K=self.N else: self.K=K self.k=self.K.bit_length() self.permutation_doubling=[[self.permutation[n]] for n in range(self.N)] if self.lst!=None: self.doubling=[[self.lst[n]] for n in range(self.N)] for k in range(1,self.k): for n in range(self.N): if self.permutation_doubling[n][k-1]==None: self.permutation_doubling[n].append(None) if self.f!=None: self.doubling[n].append(None) if self.permutation_doubling[n][k-1]!=None: self.permutation_doubling[n].append(self.permutation_doubling[self.permutation_doubling[n][k-1]][k-1]) if self.f!=None: self.doubling[n].append(self.f(self.doubling[n][k-1],self.doubling[self.permutation_doubling[n][k-1]][k-1])) def Permutation_Doubling(self,N,K): if K<0 or 1<<self.k<=K: return None for k in range(self.k): if K>>k&1 and N!=None: N=self.permutation_doubling[N][k] return N def Doubling(self,N,K): if K<0: return self.e retu=self.e for k in range(self.k): if K>>k&1: if self.permutation_doubling[N][k]==None: return None retu=self.f(retu,self.doubling[N][k]) N=self.permutation_doubling[N][k] return retu def Bisect(self,x,is_ok): if not is_ok(x): return -1,None K=0 for k in range(self.k-1,-1,-1): if K|1<<k<=self.K and is_ok(self.permutation_doubling[x][k]): K|=1<<k x=self.permutation_doubling[x][k] return K,x N,K=map(int,readline().split()) P=list(map(int,readline().split())) PD=Path_Doubling(N,[(P[i]+i)%N for i in range(N)],[(i+P[i])//N for i in range(N)],lambda x,y:x+y,0) PD.Build_Next(K) for i in range(N): pd=PD.Permutation_Doubling(i,K) d=PD.Doubling(i,K) ans=pd+d*N+1 print(ans)