P3711 倉鼠的數學題

有關伯努利數的知識可以看我的上一篇題解鏈接（寫的超詳細）。

$$\begin{gathered} F(x)=\sum _{k=0}^n{S}_k(x)a_k \\ 原本定義的S_k(x)=\sum _{i=0}^x{i}^k \\ 根據伯努利數的定義S_k^{\prime }(x)=\sum _{i=0}^{x?1}{i}^k \\ 則我們求F(x)=\sum _{k=0}^n{S}_k^{\prime }(x)a_k,答案即為F(x+1) \\ 考慮先求F(x)=\sum _{k=0}^n{a}_k\frac{1}{k+1}\sum _{i=0}^k{C}_{k+1}^i{B}_i{x}^{k?i+1} \\ \sum _{k=0}^n\frac{a_k}{k+1}\sum _{i=0}^k\frac{(k+1)!}{i!(k+1?i)!}{B}_i{x}^{k?i+1} \\ \sum _{k=0}^n{a}_{k}k!\sum _{i=0}^k\frac{B_i}{i!}\frac{x^{k?i+1}}{(k?i+1)!} \\ \sum _{i=1}^{n+1}\frac{x^i}{i!}\sum _{k=i?1}^n\left(a_{k}k!\right)\left(\frac{B_{k+1?i}}{(k+1?i)!}\right) \\ 設f(n)=a_{k}k!,g(n)=\frac{B_n}{n!},另g(n)=g(n?i)即翻轉一下 \\ \sum _{i=1}^{n+1}\frac{x^i}{i!}\sum _{k=i?1}^{n}f(k)g(n+i?k) \\ 容易發現后面是一個卷積，所以可以優化 \end{gathered}$$
下面的$n=n+1$，由上易知$F(x)$是一個$n+1$次多項式。
$$\begin{gathered} 設F(x)=\sum _{i=0}^n{a}_i{x}^i \\ F(x+1)=\sum _{i=0}^n{a}_i(x+1{)}^i \\ \sum _{i=0}^n{a}_i\sum _{j=0}^i{C}_j^i{x}^j \\ \sum _{i=0}^n{a}_i\sum _{j=0}^i\frac{i!}{j!(i?j)!}{x}^j \\ \sum _{i=0}^n\frac{x^i}{i!}\sum _{j=i}^n{a}_{j}j!\frac{1}{(j?i)!} \\ 另f(n)=a_{n}n!,g(n)=\frac{1}{n!},再翻轉g(n) \\ \sum _{i=0}^n\frac{x^i}{i!}\sum _{j=i}^{n}f(j)g(n+i?j) \\ 后面也同樣是一個卷積，所以可以優化 \end{gathered}$$

#include <bits/stdc++.h>using namespace std;const int N = 1e6 + 10, mod = 998244353;int r[N], t[N];int quick_pow(int a, int n) {int ans = 1;while (n) {if (n & 1) {ans = 1ll * a * ans % mod;}a = 1ll * a * a % mod;n >>= 1;}return ans; }void get_r(int lim) {for (int i = 0; i < lim; i++) {r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);} }void NTT(int *f, int lim, int rev) {for (int i = 0; i < lim; i++) {if (i < r[i]) {swap(f[i], f[r[i]]);}}for (int mid = 1; mid < lim; mid <<= 1) {int wn = quick_pow(3, (mod - 1) / (mid << 1));for (int len = mid << 1, cur = 0; cur < lim; cur += len) {int w = 1;for (int k = 0; k < mid; k++, w = 1ll * w * wn % mod) {int x = f[cur + k], y = 1ll * w * f[cur + mid + k] % mod;f[cur + k] = (x + y) % mod, f[cur + mid + k] = (x - y + mod) % mod;}}}if (rev == -1) {int inv = quick_pow(lim, mod - 2);reverse(f + 1, f + lim);for (int i = 0; i < lim; i++) {f[i] = 1ll * f[i] * inv % mod;}} }void polyinv(int *f, int *g, int n) {if (n == 1) {g[0] = quick_pow(f[0], mod - 2);return ;}polyinv(f, g, n + 1 >> 1);for (int i = 0; i < n; i++) {t[i] = f[i];}int lim = 1;while (lim < 2 * n) {lim <<= 1;}get_r(lim);NTT(t, lim, 1);NTT(g, lim, 1);for (int i = 0; i < lim; i++) {int cur = (2 - 1ll * g[i] * t[i] % mod + mod) % mod;g[i] = 1ll * g[i] * cur % mod;t[i] = 0;}NTT(g, lim, -1);for (int i = n; i < lim; i++) {g[i] = 0;} }int fac[N], ifac[N], a[N], f[N], g[N], B[N], n;void init() {fac[0] = 1;for (int i = 1; i < N; i++) {fac[i] = 1ll * fac[i - 1] * i % mod;}ifac[N - 1] = quick_pow(fac[N - 1], mod - 2);for (int i = N - 2; i >= 0; i--) {ifac[i] = 1ll * ifac[i + 1] * (i + 1) % mod;} }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();scanf("%d", &n);for (int i = 0; i <= n; i++) {scanf("%d", &a[i]);}for (int i = 0; i <= n + 1; i++) {B[i] = ifac[i + 1];}polyinv(B, g, n + 2);for (int i = 0; i <= n; i++) {f[i] = 1ll * a[i] * fac[i] % mod;}reverse(g, g + n + 2);int lim = 1;while (lim < 2 * n + 10) {lim <<= 1;}get_r(lim);NTT(f, lim, 1), NTT(g, lim, 1);for (int i = 0; i < lim; i++) {f[i] = 1ll * f[i] * g[i] % mod;}NTT(f, lim, -1);for (int i = 1; i <= n + 1; i++) {a[i] = 1ll * ifac[i] * f[n + i] % mod;}a[0] = 0;n++;for (int i = 0; i < lim; i++) {f[i] = g[i] = 0;}for (int i = 0; i <= n; i++) {f[i] = 1ll * a[i] * fac[i] % mod;g[i] = ifac[n - i];}NTT(f, lim, 1), NTT(g, lim, 1);for (int i = 0; i < lim; i++) {f[i] = 1ll * f[i] * g[i] % mod;}NTT(f, lim, -1);for (int i = 0; i <= n; i++) {printf("%lld%c", 1ll * ifac[i] * f[n + i] % mod, i == n ? '\n' : ' ');}return 0; }

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