結果

問題 No.2181 LRM Question 2
ユーザー MasKoaTS
提出日時 2022-10-19 20:42:54
言語 PyPy3
(7.3.15)
結果
RE  
実行時間 -
コード長 2,369 bytes
コンパイル時間 286 ms
コンパイル使用メモリ 82,048 KB
実行使用メモリ 383,200 KB
最終ジャッジ日時 2024-06-29 21:42:07
合計ジャッジ時間 9,972 ms
ジャッジサーバーID
(参考情報)
judge2 / judge5
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 1 RE * 2
other AC * 7 RE * 16
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
input = sys.stdin.readline
class BinomialCoefficient:
def __init__(self, mod):
self.mod = mod
self.prime = self.prime_factorize(mod)
self.facs = []
self.invs = []
self.pows = []
self.mods = {}
for p, c in self.prime:
pc = pow(p, c)
fac = [1] * pc
inv = [1] * pc
for i in range(1, pc):
k = i
if(i % p == 0):
k = 1
fac[i] = fac[i - 1] * k % pc
inv[-1] = fac[-1]
for i in range(pc - 1, 0, -1):
k = i
if(i % p == 0):
k = 1
inv[i - 1] = inv[i] * k % pc
self.facs.append(fac)
self.invs.append(inv)
pw = [1]
while(pw[-1] * p != pc):
pw.append(pw[-1] * p)
self.pows.append(pw)
r = [0] * len(self.prime)
pcs = [pow(p, c) for p, c in self.prime]
for i in range(1, mod):
for j, pc in enumerate(pcs):
r[j] += 1
if(r[j] >= pc):
r[j] -= pc
self.mods[tuple(r)] = i
def prime_factorize(self, n):
prime = []
f = 2
while(f * f <= n):
if(n % f == 0):
n //= f
cnt = 1
while(n % f == 0):
n //= f
cnt += 1
prime.append((f, cnt))
f += 1
if(n != 1):
prime.append((n, 1))
return prime
def inv_gcd(self, n, m):
n %= m
if(n == 0):
return m, 0
s, t, m0, m1 = m, n, 0, 1
while(t):
u = s // t
s -= t * u
m0 -= m1 * u
m0, m1, s, t = m1, m0, t, s
if(m0 < 0):
m0 += m // s
return s, m0
def inv_mod(self, n, m):
g, im = self.inv_gcd(n, m)
return im
def calc_e(self, n, k, r, p):
e = 0
while(n):
n //= p
e += n
while(k):
k //= p
e -= k
while(r):
r //= p
e -= r
return e
def lucas(self, n, k, p, c, i):
pw = self.pows[i]
fac = self.facs[i]
inv = self.invs[i]
r = n - k
pc = pow(p, c)
e = self.calc_e(n, k, r, p)
if(e >= len(pw)):
return 0
ret = pw[e]
if((p != 2 or c < 3) and (self.calc_e(n // pw[-1], k // pw[-1], r // pw[-1], p) & 1)):
ret *= -1
while(n):
ret *= fac[n % pc] * inv[k % pc] * inv[r % pc]
ret %= pc
n //= p
k //= p
r //= p
return ret
def __call__(self, n, k):
if(k < 0 or k > n):
return 0
if(k == 0 or k == n):
return 1
r = [self.lucas(n, k, p, c, i) for i, (p, c) in enumerate(self.prime)]
return self.mods[tuple(r)]
"""
Main Code
"""
l, r, m = map(int, input().split())
nCk = BinomialCoefficient(m)
ans = 0
for i in range(l, r + 1):
ans += nCk(2 * i, i) + m - 2
ans %= m
print(ans)
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