結果

問題 No.1547 [Cherry 2nd Tune *] 偶然の勝利の確率
ユーザー mkawa2
提出日時 2021-06-11 23:33:00
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 1,538 ms / 2,000 ms
コード長 1,959 bytes
コンパイル時間 308 ms
コンパイル使用メモリ 82,596 KB
実行使用メモリ 83,304 KB
最終ジャッジ日時 2024-12-15 03:07:18
合計ジャッジ時間 22,017 ms
ジャッジサーバーID
(参考情報)
judge4 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
other AC * 36
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

import sys
sys.setrecursionlimit(200005)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LI1(): return list(map(int1, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
inf = 10**16
md = 998244353
# md = 10**9+7
def dot(aa,bb):
return [[sum(a*b%md for a,b in zip(row,col))%md for col in zip(*bb)] for row in aa]
ma, na, s = LI()
mb, nb, t = LI()
K = II()
s += t
s, t = t, s
pa = ma*pow(na, md-2, md)
pb = mb*pow(nb, md-2, md)
# print(pa,pb)
dtop = [0]*(t+1)
now = (1-pa)%md
for i in range(t+1):
dtop[i] = now
now = now*pa%md
pp0=[[0]*(t+1) for _ in range(t+1)]
for i in range(1,t):
p=1
d=0
for j in range(i,t):
pp0[i][j]=dtop[d]
p-=dtop[d]
d+=1
pp0[i][t]=p%md
# p2D(pp0)
dtop = [0]*(t+1)
now = (1-pb)%md
for i in range(t+1):
dtop[i] = now
now = now*pb%md
pp1=[[0]*(t+1) for _ in range(t+1)]
for i in range(t-1,0,-1):
p=1
d=0
for j in range(i,0,-1):
pp1[i][j]=dtop[d]
p-=dtop[d]
d+=1
pp1[i][0]=p%md
base=[[0]*(t+1) for _ in range(t+1)]
for i in range(1,t):
for j in range(i,t):
for k in range(j+1):
base[k][i]+=pp0[i][j]*pp1[j][k]
base[k][i]%=md
for i in range(1,t):
base[t][i]=pp0[i][t]
base[0][0]=1
base[t][t]=1
# p2D(base)
cc=[[0]*(t+1) for _ in range(t+1)]
for i in range(t+1):cc[i][i]=1
while K:
if K&1:cc=dot(cc,base)
base=dot(base,base)
K>>=1
# print(s,t)
# p2D(cc)
print(cc[t][s])
print(cc[0][s])
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