51nod 1514 美妙的序列
如果一个长度为 $n$ 的排列的任意一个严格前缀都不是一个 $1 \sim i$ 的排列，则称这个排列是合法的，$T$ 次求有多少长度为 $n$ 的合法的排列
$T,n \leq 10^5$
sol：
小清新多项式题，考虑 dp，$f_i$ 表示长度为 $i$ 的合法排列数量，$g_i$ 表示长度为 $i$ 的所有排列数量，考虑容斥
一个不合法的排列一定存在至少一个 $x$ 使 $[1,x]$ 是排列，$[x+1,n]$ 是排列
但因为是“至少一个”没法枚举，但发现我们可以只枚举第一个 $x$ 的位置，后面随便排就好了，因为此时已经不合法
枚举 $x$ 相当于枚举当前排列中极长的合法前缀长度，有 $g_n = \sum\limits_{i=0}^{n-1} f_i \times g_{n-i}$
然后发现是个多项式求逆的标准形式
如何求 $g_i$ 呢？很明显，$g_n = n!$
#include <bits/stdc++.h> #define LL long long #define rep(i, s, t) for (register int i = (s), i##end = (t); i <= i##end; ++i) #define dwn(i, s, t) for (register int i = (s), i##end = (t); i >= i##end; --i) using namespace std; inline int read() { int x = 0, f = 1; char ch; for (ch = getchar(); !isdigit(ch); ch = getchar()) if (ch == '-') f = -f; for (; isdigit(ch); ch = getchar()) x = 10 * x + ch - '0'; return x * f; } const int maxn = 600010, mod = 998244353; int F[maxn], G[maxn]; inline int skr(int x, int t) { int res = 1; for (; t; x = 1LL * x * x % mod, t >>= 1) if (t & 1) res = 1LL * res * x % mod; return res; } int r[maxn], lg[maxn], temp[maxn]; int inv[maxn], ifac[maxn], fac[maxn]; inline int skr(int x, LL t) { int res = 1; while (t) { if (t & 1) res = 1LL * res * x % mod; x = 1LL * x * x % mod; t = t >> 1; } return res; } inline void fft(int *a, int n, int type) { for (int i = 0; i < n; i++) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (lg[n] - 1)); for (int i = 0; i < n; i++) if (i < r[i]) swap(a[i], a[r[i]]); for (int i = 1; i < n; i <<= 1) { int wn = skr(3, (mod - 1) / (i << 1)); if (type == -1) wn = skr(wn, mod - 2); // cout << wn << endl; for (int j = 0; j < n; j += (i << 1)) { int w = 1; for (int k = 0; k < i; k++, w = (1LL * (LL)w * (LL)wn) % mod) { int x = a[j + k], y = (1LL * (LL)w * (LL)a[j + k + i]) % mod; a[j + k] = (x + y) % mod; a[j + k + i] = (((x - y) % mod) + mod) % mod; } } } if (type == -1) { int inv = skr(n, mod - 2); for (int i = 0; i < n; i++) a[i] = ((LL)a[i] * (LL)inv) % mod; } } inline void Inverse(int *a, int *b, int n) { if (n == 1) { b[0] = skr(a[0], mod - 2); return; } Inverse(a, b, n >> 1); memcpy(temp, a, n * sizeof(int)); memset(temp + n, 0, n * sizeof(int)); fft(temp, n << 1, 1); fft(b, n << 1, 1); for (int i = 0; i < (n << 1); i++) b[i] = 1LL * b[i] * ((2LL - 1LL * temp[i] * b[i] % mod + mod) % mod) % mod; fft(b, n << 1, -1); memset(b + n, 0, n * sizeof(int)); } int c[maxn], d[maxn]; inline void Ln(int *a, int *b, int n) { Inverse(a, c, n); for (int i = 0; i < n - 1; ++i) d[i] = (LL)(i + 1) * a[i + 1] % mod; d[n - 1] = 0; fft(c, n << 1, 1); fft(d, n << 1, 1); for (int i = 0; i < (n << 1); ++i) c[i] = 1LL * d[i] * c[i] % mod; fft(c, (n << 1), -1); for (int i = 1; i < (n << 1); ++i) b[i] = 1LL * inv[i] * c[i - 1] % mod; b[0] = 0; for (int i = 0; i < (n << 1); ++i) c[i] = d[i] = 0; } int temp_w[maxn], temp_Ln[maxn]; inline void Exp(int *a, int *b, int n) { if (n == 1) { b[0] = 1; return; } Exp(a, b, n >> 1); memcpy(temp_w, b, sizeof(int) * n); memset(temp_w + n, 0, sizeof(int) * n); Ln(b, temp_Ln, n); for (int i = 0; i < n; i++) temp_Ln[i] = (mod + a[i] - temp_Ln[i]) % mod; (temp_Ln[0] += 1) %= mod; fft(temp_w, n << 1, 1); fft(temp_Ln, n << 1, 1); for (int i = 0; i < (n << 1); i++) temp_w[i] = 1LL * temp_w[i] * temp_Ln[i] % mod; fft(temp_w, n << 1, -1); memcpy(b, temp_w, n * sizeof(int)); memset(b + n, 0, n * sizeof(int)); memset(temp_w, 0, sizeof(int) * (n << 1)); memset(temp_Ln, 0, sizeof(int) * (n << 1)); } int main() { lg[0] = -1; rep(i, 1, 600000) lg[i] = lg[i >> 1] + 1; inv[1] = ifac[0] = fac[0] = 1; rep(i, 1, 600000) { if (i != 1) inv[i] = -(LL)mod / i * inv[mod % i] % mod; inv[i] = ((inv[i] % mod) + mod) % mod; ifac[i] = (LL)ifac[i - 1] * inv[i] % mod; fac[i] = (LL)fac[i - 1] * i % mod; } int len = 1; for(; len <= 200000; len <<= 1); memcpy(F, fac, sizeof(F)); Inverse(F, G, len); rep(i, 0, len-1) G[i] = mod - G[i]; G[0]++; int T = read(); while(T--) cout << G[read()] << '\n'; }
