結果
問題 | No.1712 Read and Pile |
ユーザー |
|
提出日時 | 2021-05-10 23:39:16 |
言語 | PyPy3 (7.3.15) |
結果 |
AC
|
実行時間 | 671 ms / 2,000 ms |
コード長 | 2,823 bytes |
コンパイル時間 | 349 ms |
コンパイル使用メモリ | 82,320 KB |
実行使用メモリ | 132,008 KB |
最終ジャッジ日時 | 2024-09-17 16:56:42 |
合計ジャッジ時間 | 16,156 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 40 |
ソースコード
class BIT():def __init__(self,n,mod=None):self.BIT=[0]*(n+1)self.num=nself.mod = moddef query(self,idx):res_sum = 0while idx > 0:res_sum += self.BIT[idx]if self.mod:res_sum %= self.modidx -= idx&(-idx)return res_sum#Ai += x O(logN)def update(self,idx,x):while idx <= self.num:self.BIT[idx] += xif self.mod:self.BIT[idx] %= self.modidx += idx&(-idx)returnimport sys,random,bisectfrom collections import deque,defaultdictfrom heapq import heapify,heappop,heappushfrom itertools import permutationsfrom math import log,gcdinput = lambda :sys.stdin.readline().rstrip()mi = lambda :map(int,input().split())li = lambda :list(mi())mod = 998244353i2 = pow(2,mod-2,mod)def solve(N,M,A):assert 1<=N<=2*10**5assert 1<=M<=2*10**5for i in range(M):if A[i]!=-1:assert 1<=A[i]<=Nif N==1:return Mif N==2:res = Mpre = 1Z = A.count(-1)for i in range(M):if A[i]==-1:res += i2res %= modelse:if pre!=-1 and A[i]!=pre:res += 1res %= modelif pre==-1:res += i2res %= modpre = A[i]return resp = ((N-2)*pow(N,mod-2,mod)) % modip = pow(p,mod-2,mod)pow_p = [1 for i in range(N+M+1)]pow_ip = [1 for i in range(N+M+1)]for i in range(1,N+M+1):pow_p[i] = (pow_p[i-1] * p) % modpow_ip[i] = (pow_ip[i-1] * ip) % modA = [N-i for i in range(N)] + AA = [-100] + Azero = [A[i]==-1 for i in range(N+M+1)]for i in range(1,N+M+1):zero[i] += zero[i-1]lastappear = [N-i+1 for i in range(N+1)]res = 0bit_p = BIT(N+M,mod=998244353)bit_cnt = BIT(N+M)for i in range(1,N+1):bit_cnt.update(i,1)bit_p.update(i,pow_ip[zero[i]])for i in range(N+1,N+M+1):if A[i]==-1:res += (N-1)res %= modelse:pre = lastappear[A[i]]k = bit_cnt.query(pre-1)res = (res + N-1) % modres += (bit_p.query(i)-bit_p.query(pre)) * pow_p[zero[i]] % modres %= modres -= k * pow_p[zero[i]-zero[pre]] % modres %= modbit_cnt.update(pre,-1)bit_p.update(pre,-pow_ip[zero[pre]])lastappear[A[i]] = ibit_cnt.update(i,1)bit_p.update(i,pow_ip[zero[i]])res *= i2res %= modres += Mres %= modreturn resN,M = mi()A = li()print(solve(N,M,A))