iis网站目录权限优化方案电子版

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 % MODinv_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 % MODdef 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 % MODinv_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 % MODdef comb(n: int, k: int) -> int: # 啥时候填return fac[n] * inv_fac[k] % MOD * inv_fac[n - k] % MODclass Solution:def numberOfSequence(self, n: int, a: List[int]) -> int:m = len(a)total = n - mans = comb(total, a[0]) * comb(total - a[0], n - a[-1] - 1) % MODtotal -= a[0] + n - a[-1] - 1e = 0for p, q in pairwise(a):k = q - p - 1if 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) % MODtotal -= kreturn ans * pow(2, e, MOD) % MOD