1. 题目截图
2.题目分析
需要把其分为多个段进行填充
长为k的段，从两端往中间填充的方案数有2 ** (k - 1)种
组合数就是选哪几个数填哪几个段即可
3.组合数含逆元模版
MOD = 1_000_000_007
MX = 100_000
# 组合数模板
fac = [0] * MX
fac[0] = 1
for i in range(1, MX):
fac[i] = fac[i - 1] * i % MOD
inv_fac = [0] * MX
inv_fac[MX - 1] = pow(fac[MX - 1], -1, MOD)
for i in range(MX - 1, 0, -1):
inv_fac[i - 1] = inv_fac[i] * i % MOD
def comb(n: int, k: int) -> int: # 啥时候填
return fac[n] * inv_fac[k] % MOD * inv_fac[n - k] % MOD
ac code
MOD = 1_000_000_007
MX = 100_000
# 组合数模板
fac = [0] * MX
fac[0] = 1
for i in range(1, MX):
fac[i] = fac[i - 1] * i % MOD
inv_fac = [0] * MX
inv_fac[MX - 1] = pow(fac[MX - 1], -1, MOD)
for i in range(MX - 1, 0, -1):
inv_fac[i - 1] = inv_fac[i] * i % MOD
def comb(n: int, k: int) -> int: # 啥时候填
return fac[n] * inv_fac[k] % MOD * inv_fac[n - k] % MOD
class Solution:
def numberOfSequence(self, n: int, a: List[int]) -> int:
m = len(a)
total = n - m
ans = comb(total, a[0]) * comb(total - a[0], n - a[-1] - 1) % MOD
total -= a[0] + n - a[-1] - 1
e = 0
for p, q in pairwise(a):
k = q - p - 1
if k:
e += k - 1 # 长度为k的连续序列填满的种数有2 ** (k - 1)
ans = ans * comb(total, k) % MOD
total -= k
return ans * pow(2, e, MOD) % MOD
