單位根反演

一個等式：$[n\mid a]=\frac{1}{n}\sum _{k=0}^{n?1}{w}_n^{ak}$

證明：

$w_n^a$是$n$次單位根的$a$次方，所以這里是一個公比為$w_n^a$的等比數列，

考慮公比不為$1$，等比數列求和$\frac{1}{n}\frac{w_n^0(w_n^{an}?1)}{w_n^a?1}$， 有$w_n^{an}?1=w_1^a?1=0$，所以該等比數列求和為$0$，

如果$n\mid a$，有$w_n^{ak}=w_1^{\frac{a}{n}k}=1$，所以有$[n\mid a]=\frac{1}{n}\sum _{k=0}^{n?1}1=1$。

而有下面的推論：

$a\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}n)$，有$a?b\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}n)$，$[n\mid a?b]=\frac{1}{n}\sum _{k=0}^{n?1}{w}_n^{ka}{w}_n^{?kb}$。

#6485. LJJ 學二項式定理

$$\begin{gathered} \sum _{i=0}^n{(}_i^n)s^i{a}_{i\mathrm{m}\mathrm{o}\mathrm{d}4} \\ \sum _{i=0}^n{(}_i^n)s^i\sum _{j=0}^3{a}_j[i\equiv j(\mathrm{m}\mathrm{o}\mathrm{d}4)] \\ \sum _{i=0}^n{(}_i^n)s^i\sum _{j=0}^3{a}_j\frac{1}{4}\sum _{k=0}^3{w}_4^{k(i?j)} \\ \frac{1}{4}\sum _{j=0}^3{a}_j\sum _{k=0}^3{w}_4^{?kj}\sum _{i=0}^n{(}_i^n)s^i{w}_4^{ki} \\ \frac{1}{4}\sum _{j=0}^3{a}_j\sum _{k=0}^3{w}_4^{?kj}(s{w}_4^k+1{)}^n \end{gathered}$$
有$998244353$下的單位根是$3$，所以$w_4^1=3^{\frac{mod?1}{4}}$，$(w_4^1{)}^4=3^{mod?1}=1$，快速冪即可。

#include <bits/stdc++.h>using namespace std;const int mod = 998244353, w[4] = {1, 911660635, 998244352, 86583718};int quick_pow(int a, int n) {int ans = 1;while (n) {if (n & 1) {ans = 1ll * ans * a % mod;}a = 1ll * a * a % mod;n >>= 1;}return ans; }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);int T, inv4 = quick_pow(4, mod - 2);scanf("%d", &T);while (T--) {long long n, ans = 0, sum, s, a[10];scanf("%lld %lld %lld %lld %lld %lld", &n, &s, &a[0], &a[1], &a[2], &a[3]);n %= mod - 1;for (int j = 0; j < 4; j++) {sum = 0;for (int k = 0; k < 4; k++) {int cur = (4 - j * k % 4) % 4;sum = (sum + 1ll * w[cur] * quick_pow(1ll * s * w[k] % mod + 1, n) % mod) % mod;}ans = (ans + sum * a[j] % mod) % mod;}printf("%lld\n", ans * quick_pow(4, mod - 2) % mod);}return 0; }

P5591 小豬佩奇學數學

$$\begin{gathered} \sum _{i=0}^n{(}_i^n)\times p^i\times ?\frac{i}{k}? \\ ?\frac{i}{k}?=\frac{i?i\%k}{k} \\ \frac{1}{k}\sum _{i=0}^n{(}_i^n)\times p^i\times (i?i\%k) \\ \frac{1}{k}\sum _{i=0}^n{(}_i^n)\times p^i\times i?\frac{1}{k}\sum _{i=0}^n{(}_i^n)\times p^i(i\mathrm{m}\mathrm{o}\mathrm{d}k) \end{gathered}$$
考慮前項化簡：
$$\begin{gathered} \frac{1}{k}\sum _{i=0}^n{(}_i^n)\times p^i\times i \\ \sum _{i=0}^n{(}_i^n)\times i=\sum _{i=0}^n\frac{n!}{i!(n?i)!}i=\sum _{i=1}^n\frac{n!}{(i?1)!(n?i)!} \\ n\sum _{i=1}^n\frac{(n?1)!}{(i?1)!(n?i)!}=n\sum _{i=1}^n{(}_{i?1}^{n?1}) \\ 原式\frac{np}{k}\sum _{i=1}^n{(}_{i?1}^{n?1})p^{i?1}=\frac{np}{k}(p+1{)}^{n?1} \end{gathered}$$
考慮后項化簡：
$$\begin{gathered} \frac{1}{k}\sum _{d=0}^{k?1}d\sum _{i=0}^n{(}_i^n)p^i[k\mid i?d] \\ 對[k\mid i?d]進行單位根反演 \\ \frac{1}{k}\sum _{d=0}^{k?1}d\sum _{i=0}^n{(}_i^n)\times p^i\times \frac{1}{k}\sum _{j=0}^{k?1}{w}_k^{j(i?d)} \\ \frac{1}{k^2}\sum _{d=0}^{k?1}d\sum _{j=0}^{k?1}{w}_k^{?jd}\sum _{i=0}^n{(}_i^n)p^i{w}_k^{ij} \\ \frac{1}{k^2}\sum _{d=0}^{k?1}d\sum _{j=0}^{k?1}{w}_k^{?jd}(p{w}_k^j+1{)}^n \\ \frac{1}{k^2}\sum _{j=0}^{k?1}(p\times w_k^j+1{)}^n\sum _{d=0}^{k?1}d\times w_k^{?jd} \end{gathered}$$
$\sum _{d=0}^{k?1}d\times w_k^{?jd}$形如$\sum _{i=0}^{n?1}i{r}^i$，考慮形如$\sum _{i=0}^{n?1}(i+1)r^{i+1}$這樣的式子化簡：
$$\begin{gathered} \sum _{i=0}^{n?1}(i+1)r^{i+1}=r\sum _{i=0}^{n?1}i{r}^i+\sum _{i=0}^{n?1}{r}^{i+1}=r\sum _{i=0}^{n?1}i{r}^i+\frac{r(r^n?1)}{r?1} \\ 有\sum _{i=0}^{n?1}i{r}^i=\sum _{i=0}^{n?1}(i+1)r^{i+1}?0r^0?n{r}^n \\ 所以\sum _{i=0}^{n?1}i{r}^i=r\sum _{i=0}^{n?1}i{r}^i+\frac{r(r^n?1)}{r?1}?n{r}^n \\ 整理可得\sum _{i=0}^{n?1}i{r}^i=\frac{n{r}^n?\frac{r(r^n?1)}{r?1}}{r?1}=\frac{n{r}^n(r?1)?r(r^n?1)}{(r?1{)}^2}=\frac{n{r}^n}{r?1}?\frac{r(r^n?1)}{(r?1{)}^2} \end{gathered}$$
特殊情況$r=1$時$\sum _{i=0}^{n?1}i{r}^i=\frac{n(n?1)}{2}$，對后項再進行化簡：
$$\begin{gathered} \sum _{d=0}^{k?1}d\times w_k^{?jd} \\ 考慮j=0時，也就是w_k^j不為1 \\ \sum _{d=0}^{k?1}d\times w_k^{?jd}=\frac{k{w}_k^{?jk}}{w_k^{?j}?1}?\frac{w_k^{?j}(w_k^{?jk}?1)}{(w_k^{?j}?1{)}^2}=\frac{k}{w_k^{?j}?1} \\ 后項整體有:\frac{1}{k^2}\sum _{j=0}^{k?1}(p\times w_k^j+1{)}^n\frac{k}{w_k^{?j}?1} \end{gathered}$$
最后答案為：
$$\begin{gathered} \frac{np}{k}(p+1{)}^{n?1}?\left(\frac{k(k?1)}{2}\frac{(p+1{)}^n}{k^2}+\sum _{j=1}^{k?1}\frac{(p\times w_k^j+1{)}^n}{k(w_k^{?j}?1)}\right) \\ \frac{np}{k}(p+1{)}^{n?1}?\left(\frac{(k?1)(p+1{)}^n}{2k}+\sum _{j=1}^{k?1}\frac{(p\times w_k^j+1{)}^n}{k(w_k^{?j}?1)}\right) \end{gathered}$$

#include <bits/stdc++.h>using namespace std;const int mod = 998244353, N = 2e6 + 10;int w[N], n, p, k;int quick_pow(int a, int n) {int ans = 1;while (n) {if (n & 1) {ans = 1ll * ans * a % mod;}a = 1ll * a * a % mod;n >>= 1;}return ans; }int inv(int x) {return quick_pow(x, mod - 2); }int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);scanf("%d %d %d", &n, &p, &k);int wn = quick_pow(3, (mod - 1) / k);int ans = 1ll * n * p % mod * quick_pow(p + 1, n - 1) % mod * inv(k) % mod;int res = 1ll * (k - 1) * quick_pow(p + 1, n) % mod * inv(2 * k) % mod;w[0] = 1;for (int i = 1; i < k; i++) {w[i] = 1ll * w[i - 1] * wn % mod;}for (int i = 1; i < k; i++) {res = (res + 1ll * quick_pow(1ll * p * w[i] % mod + 1, n) * inv(1ll * k * (w[k - i] - 1) % mod) % mod) % mod;}ans = (ans - res + mod) % mod;printf("%d\n", ans);return 0; }

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