Gcd Product

$$\begin{gathered} C_m=\sum _{i=1}^m{A}_{\gcd (i,m)}{B}_{\gcd (k+1?i,m)} \\ \sum _{d1\mid m}{A}_{d1}\sum _{d_2\mid m}{B}_{d_2}\sum _{i=1}^m\left([\gcd (\frac{i}{d_1},\frac{m}{d_1})=1][d_1\mid i]\right)\left([\gcd (\frac{m+1?i}{d_2},\frac{m}{d_2})=1][d_2\mid m+1?i]\right) \\ \sum _{d_1\mid m}{A}_{d_1}\sum _{d_2\mid m}{B}_{d_2}\sum _{k_1\mid \frac{m}{d_1}}\mu (k_1)\sum _{k_2\mid \frac{m}{d_2}}\mu (k_2)\sum _{i=1}^m\left([d_1\mid i][k_1\mid \frac{i}{d_1}]\right)\left([d_2\mid m+1?i][k_2\mid \frac{m+1?i}{d_2}]\right) \\ {T}_1=d_1\times k_1,T_2=d_2\times k_2 \\ \sum _{T_1\mid m}\sum _{d_1\mid T_1}{A}_{d_1}\mu (\frac{T_1}{d_1})\sum _{T_2\mid m}\sum _{d_2\mid T_2}{B}_{d_2}\mu (\frac{T_2}{d_2})\sum _{i=1}^m[T_1\mid i][T_2\mid m+1?i] \end{gathered}$$
觀察式子，不難發(fā)現(xiàn)$\sum _{d_1\mid T_1}{A}_{d_1}\mu (\frac{T_1}{d_1}),{\sum }_{d_2\mid T_2}{B}_{d_2}\mu (\frac{T_2}{d_2})$，二者對于給定的$T_1,T_2$都是可以確定的，跟變量$m$無關(guān)，

設(shè)$f(n)=\sum _{d\mid n}{A}_d\mu (\frac{n}{d}),g(n)={\sum }_{d\mid n}{B}_d\mu (\frac{n}{d})$，得到$\sum _{T_1\mid m}f(T_1)\sum _{T_2\mid m}g(T_2)\sum _{i=1}^m[T_1\mid i][T_2\mid m+1?i]$，

接下來我們考慮化簡$\sum _{i=1}^m[T_1\mid i][T_2\mid m+1?i]$。

$設(shè)i=T_1{k}_1,m+1?i=T_2{k}_2$，則$T_1{k}_1+T_2{k}_2=m+1$，可得：

同余方程$T_1{k}_1\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{T}_2),T_2{k}_2\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}{T}_1)$。

由$T_1\mid i,T_2\mid m+1?i$，則$\gcd (T_1,T_2)\mid i,\gcd (T_1,T_2)\mid m+1?i$，所以$\gcd (T_1,T_2)\mid m+1$。

由$T_1\mid m,T_2\mid m$，則$\gcd (T_1,T_2)\mid m$，因為$gcd(m,m+1)=1$，所以有$gcd(T_1,T_2)=1$。

所以$k_1,k_2$，分別在膜$T_2,T_1$下有且只有唯一解$x_1,x_2$，有$x_1{T}_1<T_1{T}_2,x_2{T}_2<T_1{T}_2$，

可得$x_1{T}_1+x_2{T}_2=T_1{T}_2+1$，要使$k_1{T}_1+k_2{T}_2=m+1$，

相當(dāng)于在$x_1{T}_1+x_2{T}_2=T_1{T}_2+1$的基礎(chǔ)上給$x_1{T}_1,x_2{T}_2$組合分配，湊得$m?T_1{T}_2$。

設(shè)$m=K{T}_1{T}_2$，所以解的個數(shù)就是$K=\frac{m}{T_1{T}_2}$，則有：
$$\begin{gathered} C_m=\sum _{T_1\mid m}f(T_1)\sum _{T_2\mid m}g(T_2)\frac{m}{T_1{T}_2}[\gcd (T_1,T_2)=1] \\ T=T_1{T}_2 \\ \sum _{T\mid m}\frac{m}{T}\sum _{T_1\mid T}f(T_1)g(\frac{T}{T_1})[gcd(T_1,\frac{T}{T_2})=1] \end{gathered}$$
先做一次迪利克雷卷積得到$f,g$，再做一次互質(zhì)迪利克雷卷積得到$\sum _{T_1\mid T}f(T_1)g(\frac{T}{T_1})[gcd(T_1,\frac{T}{T_2})=1]$，最后迪利克雷卷積得到答案。

#include <bits/stdc++.h>using namespace std;const int N = 5e5 + 10, mod = 998244353;int prime[N], mu[N], phi[N], A[N], B[N], f[N], g[N], h[N], ans[N], n, cnt;bool st[N];void init() {mu[1] = phi[1] = 1;for (int i = 2; i < N; i++) {if (!st[i]) {prime[++cnt] = i;mu[i] = mod - 1;phi[i] = i - 1;}for (int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1;if (i % prime[j] == 0) {phi[i * prime[j]] = phi[i] * prime[j];break;}phi[i * prime[j]] = phi[i] * (prime[j] - 1);mu[i * prime[j]] = mod - mu[i];}} }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 = 1; i <= n; i++) {scanf("%d", &A[i]);}for (int i = 1; i <= n; i++) {scanf("%d", &B[i]);}for (int i = 1; i <= n; i++) {for (int j = i; j <= n; j += i) {f[j] = (f[j] + 1ll * A[i] * mu[j / i] % mod) % mod;g[j] = (g[j] + 1ll * B[i] * mu[j / i] % mod) % mod;}}for (int i = 1; i <= n; i++) {for (int j = i; j <= n; j += i) {if (1ll * phi[i] * phi[j / i] == phi[j]) {h[j] = (h[j] + 1ll * f[i] * g[j / i] % mod) % mod;}}}for (int i = 1; i <= n; i++) {for (int j = i; j <= n; j += i) {ans[j] = (ans[j] + 1ll * (j / i) * h[i] % mod) % mod;}}for (int i = 1; i <= n; i++) {ans[i] ^= ans[i - 1];}printf("%d\n", ans[n]);return 0; }

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