結果

問題 No.1547 [Cherry 2nd Tune *] 偶然の勝利の確率
コンテスト
ユーザー mkawa2
提出日時 2021-06-11 23:40:26
言語 PyPy3
(7.3.15)
結果
AC  
実行時間 1,549 ms / 2,000 ms
コード長 2,043 bytes
コンパイル時間 460 ms
コンパイル使用メモリ 82,536 KB
実行使用メモリ 83,120 KB
最終ジャッジ日時 2024-12-15 03:23:24
合計ジャッジ時間 22,133 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge4
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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 # 

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

for i in range(K.bit_length()):
    if K >> i & 1: cc = dot(cc, base)
    base = dot(base, base)

# print(s,t)
# p2D(cc)
print(cc[t][s])
print(cc[0][s])
0