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P5591 小猪佩奇学数学（单位根反演）

發(fā)布時(shí)間：2023/12/4 编程问答 33 豆豆

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P5591 小豬佩奇學(xué)數(shù)學(xué)

$∑i=0n(in)×pi×?ik??ik?=i?i%kk1k∑i=0n(in)×pi×(i?i%k)1k∑i=0n(in)×pi×i?1k∑i=0n(in)×pi(imodk)\sum_{i = 0} ^{n} (_i ^ n) \times p ^ i \times \lfloor \frac{i}{k} \rfloor\\ \lfloor \frac{i}{k} \rfloor = \frac{i - i \% k}{k}\\ \frac{1}{k} \sum_{i = 0} ^{n} (_i ^ n) \times p ^ i \times (i - i \% k)\\ \frac{1}{k} \sum_{i = 0} ^{n} (_i ^ n) \times p ^ i \times i - \frac{1}{k} \sum_{i = 0} ^{n} (_i ^ n) \times p ^ i (i \mod k)\\$
考慮前項(xiàng)化簡(jiǎn)：
$1k∑i=0n(in)×pi×i∑i=0n(in)×i=∑i=0nn!i!(n?i)!i=∑i=1nn!(i?1)!(n?i)!n∑i=1n(n?1)!(i?1)!(n?i)!=n∑i=1n(i?1n?1)原式npk∑i=1n(i?1n?1)pi?1=npk(p+1)n?1\frac{1}{k} \sum_{i = 0} ^{n} (_i ^ n) \times p ^ i \times i\\ \sum_{i = 0} ^{n} (_i ^ n) \times i = \sum_{i = 0} ^{n} \frac{n!}{i!(n - i)!} i = \sum_{i = 1} ^{n} \frac{n!}{(i - 1)!(n - i)!}\\ n\sum_{i = 1} ^{n} \frac{(n - 1)!}{(i - 1)!(n - i)!} = n \sum_{i = 1} ^{n} (_{i - 1} ^{n - 1})\\ 原式\frac{np}{k} \sum_{i = 1} ^{n} (_{i - 1} ^{n - 1}) p ^{i - 1} = \frac{np}{k} (p + 1) ^{n - 1}\\$
考慮后項(xiàng)化簡(jiǎn)：
$1k∑d=0k?1d∑i=0n(in)pi[k∣i?d]對(duì)[k∣i?d]進(jìn)行單位根反演1k∑d=0k?1d∑i=0n(in)×pi×1k∑j=0k?1wkj(i?d)1k2∑d=0k?1d∑j=0k?1wk?jd∑i=0n(in)piwkij1k2∑d=0k?1d∑j=0k?1wk?jd(pwkj+1)n1k2∑j=0k?1(p×wkj+1)n∑d=0k?1d×wk?jd\frac{1}{k} \sum_{d = 0} ^{k - 1} d \sum_{i = 0} ^{n} (_i ^ n) p ^ i[k \mid i - d]\\ 對(duì)[k \mid i - d]進(jìn)行單位根反演\\ \frac{1}{k} \sum_{d = 0} ^{k - 1} d \sum_{i = 0} ^{n} (_i ^ n) \times p ^ i \times \frac{1}{k} \sum_{j = 0} ^{k - 1} w_k ^{j(i- d)}\\ \frac{1}{k ^ 2}\sum_{d = 0} ^{k - 1} d \sum_{j = 0} ^{k - 1} w_{k} ^{-jd} \sum_{i = 0} ^{n} (_i ^ n) p ^ i w_{k} ^{ij}\\ \frac{1}{k ^ 2} \sum_{d = 0} ^{k - 1} d \sum_{j = 0} ^{k - 1} w_{k} ^{-jd} (pw_k ^{j} + 1) ^{n}\\ \frac{1}{k ^ 2} \sum_{j = 0} ^{k - 1} (p \times w_k ^ j + 1) ^n \sum_{d = 0} ^{k - 1} d \times w_{k} ^{-jd}\\$
$∑d=0k?1d×wk?jd\sum\limits_{d = 0} ^{k - 1} d \times w_k ^{-jd}$ 形如 $∑i=0n?1iri\sum\limits_{i = 0} ^{n - 1} i r ^ i$ ，考慮形如 $∑i=0n?1(i+1)ri+1\sum\limits_{i = 0} ^{n - 1} (i + 1) r ^ {i + 1}$ 這樣的式子化簡(jiǎn)：
$∑i=0n?1(i+1)ri+1=r∑i=0n?1iri+∑i=0n?1ri+1=r∑i=0n?1iri+r(rn?1)r?1有∑i=0n?1iri=∑i=0n?1(i+1)ri+1?0r0?nrn所以∑i=0n?1iri=r∑i=0n?1iri+r(rn?1)r?1?nrn整理可得∑i=0n?1iri=nrn?r(rn?1)r?1r?1=nrn(r?1)?r(rn?1)(r?1)2=nrnr?1?r(rn?1)(r?1)2\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\(zhòng)sum_{i = 0} ^{n - 1} i r ^ i = \sum_{i = 0} ^{n - 1} (i + 1) r ^{i + 1} - 0 r ^ 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{nr ^ n}{ r - 1} - \frac{r(r ^ n - 1)}{(r - 1) ^ 2}\\$
特殊情況 $r = 1$ 時(shí) $∑i=0n?1iri=n(n?1)2\sum\limits_{i = 0} ^{n - 1} i r ^ i = \frac{n(n - 1)}{2}$ ，對(duì)后項(xiàng)再進(jìn)行化簡(jiǎn)：
$∑d=0k?1d×wk?jd考慮j≠0時(shí)，也就是wkj不為1∑d=0k?1d×wk?jd=kwk?jkwk?j?1?wk?j(wk?jk?1)(wk?j?1)2=kwk?j?1后項(xiàng)整體有:1k2∑j=0k?1(p×wkj+1)nkwk?j?1\sum_{d = 0} ^{k - 1} d \times w _k ^{-jd}\\ 考慮j \neq 0時(shí)，也就是w_{k} ^{j}不為1\\ \sum_{d = 0} ^{k - 1} d \times w_k ^{-jd} = \frac{k w_k ^ {-jk}}{w_k ^{-j} - 1} - \frac{w_k ^{-j}(w_k ^{-jk} - 1)}{(w_k ^{-j} - 1) ^ 2} = \frac{k}{w_k ^{-j} - 1}\\ 后項(xiàng)整體有:\frac{1}{k ^ 2} \sum_{j = 0} ^{k - 1} (p \times w_k ^ j + 1) ^ n \frac{k}{w_k ^{-j} - 1}\\$
最后答案為：
$npk(p+1)n?1?(k(k?1)2(p+1)nk2+∑j=1k?1(p×wkj+1)nk(wk?j?1))npk(p+1)n?1?((k?1)(p+1)n2k+∑j=1k?1(p×wkj+1)nk(wk?j?1))\frac{np}{k}(p + 1) ^ {n - 1} - \left(\frac{k(k - 1)}{2}\frac{(p + 1) ^ n}{k ^ 2} + \sum\limits_{j = 1} ^{k - 1} \frac{(p \times w_k ^ j + 1) ^ n}{k(w_{k} ^{-j} - 1)}\right)\\ \frac{np}{k}(p + 1) ^ {n - 1} - \left(\frac{(k - 1)(p + 1) ^ n}{2k} + \sum\limits_{j = 1} ^{k - 1} \frac{(p \times w_k ^ j + 1) ^ n}{k(w_{k} ^{-j} - 1)}\right)\\$

#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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