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写像十二相.py
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写像十二相.py
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class 写像十二相:
"""https://qiita.com/drken/items/f2ea4b58b0d21621bd51"""
__slots__ = ("_fac", "_ifac", "_inv", "_mod")
def __init__(self, size: int, mod: int) -> None:
self._mod = mod
self._fac = [1]
self._ifac = [1]
self._inv = [1]
self._expand(size)
def query(
self,
n: int,
k: int,
*,
isBallSame: bool,
isBoxSame: bool,
atMostOneBallPerBox=False,
noLimitWithBox=False,
atLeastOneBallPerBox=False,
) -> int:
"""n个球放入k个盒子的方案数.
Args:
isBallSame (bool): 球是否有区别.
isBoxSame (bool): 盒子是否有区别.
atMostOneBalPerBox (bool, optional): 每个盒子最多放一个球.
noLimitWithBox (bool, optional): 每个盒子可以放任意个球.
atLeastOneBallPerBox (bool, optional): 每个盒子至少放一个球.
"""
limits = (atMostOneBallPerBox, noLimitWithBox, atLeastOneBallPerBox)
assert limits.count(True) == 1, "Must have one limit and only one limit with box."
if isBallSame and isBoxSame:
if atMostOneBallPerBox:
return self._solve1(n, k)
if noLimitWithBox:
return self._solve2(n, k)
if atLeastOneBallPerBox:
return self._solve3(n, k)
if not isBallSame and isBoxSame:
if atMostOneBallPerBox:
return self._solve4(n, k)
if noLimitWithBox:
return self._solve5(n, k)
if atLeastOneBallPerBox:
return self._solve6(n, k)
if isBallSame and not isBoxSame:
if atMostOneBallPerBox:
return self._solve7(n, k)
if noLimitWithBox:
return self._solve8(n, k)
if atLeastOneBallPerBox:
return self._solve9(n, k)
if not isBallSame and not isBoxSame:
if atMostOneBallPerBox:
return self._solve10(n, k)
if noLimitWithBox:
return self._solve11(n, k)
if atLeastOneBallPerBox:
return self._solve12(n, k)
raise Exception("Unreachable code.")
def _solve1(self, n: int, k: int) -> int:
"""有区别的球放入有区别的盒子(每个盒子最多放一个球)."""
return self.P(n, k)
def _solve2(self, n: int, k: int) -> int:
"""有区别的球放入有区别的盒子(每个盒子可以放任意个球)."""
return pow(k, n, self._mod)
def _solve3(self, n: int, k: int) -> int:
"""有区别的球放入有区别的盒子(每个盒子至少放一个球).
容斥原理:用总方案数减去不合法的方案数.
O(k*logn)
"""
mod = self._mod
res = 0
for i in range(k + 1):
if (k - i) & 1:
res -= self.C(k, i) * pow(i, n, mod)
else:
res += self.C(k, i) * pow(i, n, mod)
res %= mod
return res
def _solve4(self, n: int, k: int) -> int:
"""无区别的球放入有区别的盒子(每个盒子最多放一个球)."""
return self.C(n, k)
def _solve5(self, n: int, k: int) -> int:
"""无区别的球放入有区别的盒子(每个盒子可以放任意个球)."""
return self.C(n + k - 1, n)
def _solve6(self, n: int, k: int) -> int:
"""无区别的球放入有区别的盒子(每个盒子至少放一个球)."""
return self.C(n - 1, k - 1)
def _solve7(self, n: int, k: int) -> int:
"""有区别的球放入无区别的盒子(每个盒子最多放一个球)."""
return 0 if n > k else 1
def _solve8(self, n: int, k: int) -> int:
"""有区别的球放入无区别的盒子(每个盒子可以放任意个球).
贝尔数B(n,k).
O(min(n,k)*logn).
"""
return self.bell(n, k)
def _solve9(self, n: int, k: int) -> int:
"""有区别的球放入无区别的盒子(每个盒子至少放一个球).
第二类斯特林数S(n,k).
O(k*logn).
"""
return self.stirling2(n, k)
def _solve10(self, n: int, k: int) -> int:
"""无区别的球放入无区别的盒子(每个盒子最多放一个球)."""
return 0 if n > k else 1
def _solve11(self, n: int, k: int) -> int:
"""无区别的球放入无区别的盒子(每个盒子可以放任意个球).
分割数P(n,k).
"""
return self.partition(n, k)
def _solve12(self, n: int, k: int) -> int:
"""无区别的球放入无区别的盒子(每个盒子至少放一个球).
分割数P(n-k,k).
"""
if n < k:
return 0
return self.partition(n - k, k)
def fac(self, k: int) -> int:
self._expand(k)
return self._fac[k]
def ifac(self, k: int) -> int:
self._expand(k)
return self._ifac[k]
def inv(self, k: int) -> int:
self._expand(k)
return self._inv[k]
def C(self, n: int, k: int) -> int:
if n < 0 or k < 0 or n < k:
return 0
mod = self._mod
return self.fac(n) * self.ifac(k) % mod * self.ifac(n - k) % mod
def P(self, n: int, k: int) -> int:
if n < 0 or k < 0 or n < k:
return 0
mod = self._mod
return self.fac(n) * self.ifac(n - k) % mod
def H(self, n: int, k: int) -> int:
"""可重复选取元素的组合数"""
if n == 0:
return 1 if k == 0 else 0
return self.C(n + k - 1, k)
def put(self, n: int, k: int) -> int:
"""n个相同的球放入k个不同的盒子(盒子可放任意个球)的方案数."""
return self.C(n + k - 1, n)
def partition(self, n: int, k: int) -> int:
"""O(n*k)"""
dp = [[0] * (k + 1) for _ in range(n + 1)]
dp[0][0] = 1
for i in range(n + 1):
for j in range(1, k + 1):
if i >= j:
dp[i][j] = dp[i][j - 1] + dp[i - j][j]
else:
dp[i][j] = dp[i][j - 1]
return dp[n][k]
def bell(self, n: int, k: int) -> int:
"""O(min(n,k)*logn)"""
if k > n:
k = n
mod = self._mod
jsum = [0] * (k + 2)
for j in range(k + 1):
add = self.ifac(j)
if j & 1:
jsum[j + 1] = (jsum[j] - add) % mod
else:
jsum[j + 1] = (jsum[j] + add) % mod
res = 0
for i in range(k + 1):
res += pow(i, n, mod) * self.ifac(i) % MOD * jsum[k - i + 1]
res %= mod
return res
def stirling2(self, n: int, k: int) -> int:
"""O(k*logn)"""
mod = self._mod
res = 0
for i in range(k + 1):
if (k - i) & 1:
res -= self.C(k, i) * pow(i, n, mod)
else:
res += self.C(k, i) * pow(i, n, mod)
res %= mod
return res * self.ifac(k) % mod
def _expand(self, size: int) -> None:
if len(self._fac) < size + 1:
mod = self._mod
preSize = len(self._fac)
diff = size + 1 - preSize
self._fac += [1] * diff
self._ifac += [1] * diff
self._inv += [1] * diff
for i in range(preSize, size + 1):
self._fac[i] = self._fac[i - 1] * i % mod
self._ifac[size] = pow(self._fac[size], mod - 2, mod) # !modInv
for i in range(size - 1, preSize - 1, -1):
self._ifac[i] = self._ifac[i + 1] * (i + 1) % mod
for i in range(preSize, size + 1):
self._inv[i] = self._ifac[i] * self._fac[i - 1] % mod
MOD = int(1e9 + 7)
X = 写像十二相(int(1e5), MOD)
if __name__ == "__main__":
assert (X.partition(10, 5)) == 30
assert (X.stirling2(8, 2)) == 127
assert (X.bell(5, 5)) == 52
n, k = map(int, input().split())
print(X.bell(n, k))