定义$a\oplus b=\frac{a\times b}{\gcd (a,b{)}^2}$，$b_m=\sum _{i=1}^n\sum _{j=1}^n{a}_i\times j^c[i\oplus j=m]$，我们要求$b_1\oplus b_2\oplus \cdots \oplus b_n$。

因为$a\oplus b$的性质，我们可以考虑枚举$\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}$，设其为$a^{\prime },b^{\prime }$，则有$\gcd (a^{\prime },b^{\prime })$互质，
$$\begin{gathered} b_m=\sum _{i=1}^n\sum _{j=1}^n[i\times j=m\gcd (i,j)=1]\sum _{d=1}^{min(\frac{n}{i},\frac{n}{j})}(a_{id}(jd{)}^c) \\ \sum _{i=1}^n\sum _{j=1}^n[i\times j=m\gcd (i,j)=1]j^c\sum _{d=1}^{min(\frac{n}{i},\frac{n}{j})}{a}_{id}{d}^c \\ 设f(i,n)=\sum _{d=1}^n{a}_{id}{d}^c \end{gathered}$$
容易发现$min(\frac{n}{i},\frac{n}{j})$，如果考虑枚举$i$，则可以直接处理出$f(i,n)$，而且整体复杂度还是不变的。

#include <bits/stdc++.h>using namespace std;const int N = 1e6 + 10, mod = 998244353;int prime[N], phi[N], f[N], ans[N], a[N], p[N], n, c, cnt;bool st[N];inline int add(int x, int y) {return x + y < mod ? x + y : x + y - mod;
}inline int sub(int x, int y) {return x >= y ? x - y : x - y + mod;
}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;
}void init() {phi[1] = p[1] = 1;for (int i = 2; i < N; i++) {if (!st[i]) {prime[++cnt] = i;phi[i] = i - 1;p[i] = quick_pow(i, c);}for (int j = 1; j <= cnt && 1ll * i * prime[j] < N; j++) {st[i * prime[j]] = 1, p[i * prime[j]] = 1ll * p[i] * p[prime[j]] % mod;if (i % prime[j] == 0) {phi[i * prime[j]] = phi[i] * prime[j];break;}phi[i * prime[j]] = phi[i] * (prime[j] - 1);}}
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);scanf("%d %d", &n, &c);for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);}init();for (int i = 1; i <= n; i++) {for (int j = 1; i * j <= n; j++) {f[j] = add(f[j - 1], 1ll * a[i * j] * p[j] % mod);}for (int j = 1; i * j <= n; j++) {if (phi[i] * phi[j] == phi[i * j]) {ans[i * j] = add(ans[i * j], 1ll * p[j] * f[min(n / i, n / j)] % mod);}}}int res = 0;for (int i = 1; i <= n; i++) {res ^= ans[i];}printf("%d\n", res);return 0;
}