結果
| 問題 |
No.3038 シャッフルの再現
|
| ユーザー |
gew1fw
|
| 提出日時 | 2025-06-12 13:47:42 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
RE
|
| 実行時間 | - |
| コード長 | 1,977 bytes |
| コンパイル時間 | 188 ms |
| コンパイル使用メモリ | 82,256 KB |
| 実行使用メモリ | 67,836 KB |
| 最終ジャッジ日時 | 2025-06-12 13:48:19 |
| 合計ジャッジ時間 | 2,364 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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| ファイルパターン | 結果 |
|---|---|
| sample | RE * 1 |
| other | RE * 21 |
ソースコード
import sys
from math import gcd
from functools import reduce
MOD = 10**9 + 7
def factorize(m):
factors = []
i = 2
while i * i <= m:
if m % i == 0:
cnt = 0
while m % i == 0:
cnt += 1
m = m // i
factors.append((i, cnt))
i += 1
if m > 1:
factors.append((m, 1))
return factors
def generate_divisors(factors):
divisors = [1]
for (p, exp) in factors:
temp = []
for d in divisors:
current = d
for _ in range(exp + 1):
temp.append(current)
current *= p
divisors = temp
divisors = sorted(divisors)
return divisors
def fast_doubling(n, mod):
if n == 0:
return (0, 1)
a, b = fast_doubling(n >> 1, mod)
c = (a * (2 * b - a)) % mod
d = (a * a + b * b) % mod
if n & 1:
return (d, (c + d) % mod)
else:
return (c, d)
def get_pisano_period(p):
if p == 5:
return 20
mod5 = p % 5
if mod5 in (1, 4):
m = p - 1
else:
m = 2 * (p + 1)
factors = factorize(m)
divisors = generate_divisors(factors)
for d in divisors:
if d == 0:
continue
fn, fn1 = fast_doubling(d, p)
if fn % p == 0 and fn1 % p == 1:
return d
return m # Fallback, though theoretically should not reach here
def lcm(a, b):
return a * b // gcd(a, b)
def main():
input = sys.stdin.read().split()
idx = 0
N = int(input[idx])
idx +=1
periods = []
for _ in range(N):
p = int(input[idx])
k = int(input[idx+1])
idx +=2
base_period = get_pisano_period(p)
period = base_period * (p ** (k-1))
periods.append(period)
# Compute LCM of all periods
if not periods:
print(0)
return
total_lcm = reduce(lcm, periods)
print(total_lcm % MOD)
if __name__ == '__main__':
main()
gew1fw