大概就是照着题解抄了一遍吧，这道题太神仙了……

$$\begin{gathered} a_i=i^k,s_i=\sum _{j=1}^i{a}_j \\ calc\sum _{i=1}^n{s}_i[\gcd (i,n)=1] \\ \sum _{d\mid n}\mu (d)\sum _{i=1}^{\frac{n}{d}}{s}_{id} \end{gathered}$$

伯努利数对自然幂次求和有：
$$\begin{gathered} S_k(n)=\sum _{i=0}^n{i}^k=\frac{1}{k+1}\sum _{i=0}^k{C}_{k+1}^i{B}_i(n+1{)}^{k+i-1} \\ 不难发现其为一个k+1阶多项式，设S_k(x)=\sum _{i=0}^{k+1}{a}_i{x}^i \\ {S}_k(n)=\frac{1}{k+1}\sum _{i=1}^{k+1}{C}_{k+1}^i{B}_{k+1-i}(n+1{)}^i \\ \frac{1}{k+1}\sum _{i=0}^{k+1}{C}_{k+1}^i{B}_{k+1-i}(n+1{)}^i-\frac{1}{k+1}{B}_{k+1} \\ \frac{1}{k+1}\sum _{i=0}^{k+1}{C}_{k+1}^i{B}_{k+1-i}\sum _{j=0}^i{C}_i^j{n}^j-\frac{1}{k+1}{B}_{k+1} \\ \frac{1}{k+1}\sum _{j=0}^{k+1}{n}^j\sum _{i=j}^{k+1}{C}_{k+1}^i{B}_{k+1-i}{C}_i^j-\frac{1}{k+1}{B}_{k+1} \end{gathered}$$
有$a_0=0,j\ge 1$时有：
$$\begin{gathered} \frac{1}{k+1}\sum _{i=j}^{k+1}{C}_{k+1}^i{B}_{k+1-i}{C}_i^j \\ \frac{1}{k+1}\sum _{i=0}^{k+1-j}{c}_{k+1}^i{B}_i{C}_{k+1-i}^j \\ \frac{1}{k+1}\sum _{i=0}^{k+1-j}\frac{(k+1)!}{i!(k+1-i)!}\frac{(k+1-i)!}{j!(k+1-i-j)!}{B}_i \\ \frac{1}{k+1}\sum _{i=0}^{k+1-j}\frac{(k+1)!(k+1-j)!}{i!(k+1-j)!j!(k+1-i-j)!}{B}_i \\ \frac{1}{k+1}\sum _{i=0}^{k+1-j}{C}_{k+1-j}^i{C}_{k+1}^j{B}_i \\ \frac{1}{k+1}{C}_{k+1}^j\sum _{i=0}^{k+1-j}{C}_{k+1-j}^i{B}_i \\ \frac{1}{k+1}{C}_{k+1}^j\left(\sum _{i=0}^{k-j}{C}_{k-j+1}^i{B}_i+B_{k+1-j}\right) \\ 有\sum _{i=0}^k{C}_{k+1}^i{B}_i=0,k\ge 1 \\ 上式子为\frac{1}{k+1}{C}_{k+1}^j{B}_{k+1-j} \end{gathered}$$
综上有：
$$\{\begin{array}{rlr}{a}_0& =0& \\ a_i& =\frac{1}{k+1}{C}_{k+1}^i{B}_{k+1-i}& 1\le i<k\\ a_k& =B_1+B_0& \\ a_{k+1}& =\frac{1}{k+1}& \end{array}$$

再考虑对原式化简：

$a_{k,j}$为$k$次自然数幂之和的多项式的第$j$项，也就是上面化简的。
$$\begin{gathered} \sum _{d\mid n}\mu (d)\sum _{i=1}^{\frac{n}{d}}{s}_{id} \\ \sum _{d\mid n}\mu (d)\sum _{i=1}^{\frac{n}{d}}\sum _{j=0}^{k+1}{a}_{k,j}{i}^j{d}^j \\ \sum _{j=0}^{k+1}{a}_{k,j}\sum _{d\mid n}\mu (d)d^j\sum _{i=1}^{\frac{n}{d}}{i}^j \\ \sum _{j=0}^{k+1}{a}_{k,j}\sum _{d\mid n}\mu (d)d^j\sum _{i=0}^{j+1}{a}_{j,i}{n}^i{d}^{-i} \\ \sum _{d\mid n}\mu (d)\sum _{i=-1}^{k+1}{d}^i\sum _{x=0}^{k+1}\sum _{y=0}^{k+1}[x-y=i]a_{k,x}{a}_{x,y}{n}^y \\ \sum _{d\mid n}\mu (d)\sum _{i=-1}^{k+1}{d}^i{F}_{i+1} \\ \sum _{i=-1}^{k+1}{F}_{i+1}\sum _{d\mid n}\mu (d)d^i \end{gathered}$$

对$f(n)=\sum _{d\mid n}\mu (d)d^i$，考虑其积性$f(1)=1,f(p)=\mu (1)+\mu (p)p^i=1-p^i,f(p^k)=f(p)$，

所以答案为$\sum _{i=-1}^{k+1}{F}_{i+1}\sum _{d\mid n}\mu (d)d^i$，考虑如何求解$F_{i+1}$。
$$\begin{gathered} F_i=\sum _{x=0}^{k+1}\sum _{y=0}^{k+1}[x-y=i-1]a_{k,x}{a}_{x,y}{n}^y \\ \sum _{x=0}^{k+1}{a}_{k,x}{a}_{x,x-i+1}{n}^{x-i+1} \\ \sum _{x=1}^{k+1}{a}_{k,x}{a}_{x,x-i+1}{n}^{x-i+1} \\ \sum _{x=0}^k{a}_{k,x+1},a_{x+1,x-i+2}{n}^{x-i+2} \\ \sum _{x=0}^k\frac{1}{k+1}{C}_{k+1}^{x+1}{B}_{k-x}\frac{1}{x+2}{C}_{x+2}^i{B}_i{n}^{x-i+2} \\ \sum _{x=0}^k\frac{1}{k+1}\frac{(k+1)!}{(x+1)!(k-x)!}{B}_{k-x}\frac{1}{x+2}\frac{(x+2)!}{i!(x+2-i)!}{B}_i{n}^{x-i+2} \\ \frac{k!B_i}{i!}\sum _{x=0}^k\frac{B_{k-x}}{(k-x)!}\frac{n^{x-i+2}}{(x-i+2)!} \end{gathered}$$

由此我们得到一个卷积得形式，可$O(k\log k)$求解。

#include <bits/stdc++.h>using namespace std;const int N = 1e6 + 10, mod = 998244353;int r[N], t[N], B[N], b[N], fac[N], ifac[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;}
}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 n = 1e5 + 10, len = 1;for (int i = 0; i < n; i++) {B[i] = ifac[i + 1];}polyinv(B, b, n);for (int i = 0; i < n; i++) {B[i] = 1ll * b[i] * fac[i] % mod;b[i] = 0;}B[1] = mod - B[1];
}int k, m, n, a[N], num[N], c[N], d[N], pw[N];void polymult(int *a, int n, int *b, int m) {int lim = 1;while (lim < n + m) {lim <<= 1;}get_r(lim);for (int i = n; i < lim; i++) {a[i] = 0;}for (int i = m; i < lim; i++) {b[i] = 0;}NTT(a, lim, 1), NTT(b, lim, 1);for (int i = 0; i < lim; i++) {a[i] = 1ll * a[i] * b[i] % mod;}NTT(a, lim, -1);
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);init();int T;scanf("%d", &T);while (T--) {scanf("%d %d", &k, &m);n = 1;for (int i = 1; i <= m; i++) {scanf("%d %d", &a[i], &num[i]);n = 1ll * n * quick_pow(a[i], num[i]) % mod;}int lim = 1;while (lim < k * 2 + 10) {lim <<= 1;}for (int i = 0; i <= lim; i++) {c[i] = d[i] = 0;}for (int i = 0; i <= k; i++) {c[k + 1 - i] = 1ll * fac[k] * ifac[i] % mod * B[i] % mod;}int pow = 1;for (int i = 1; i <= k + 2; i++) {pow = 1ll * pow * n % mod;d[k + 2 - i] = 1ll * pow * ifac[i] % mod;}polymult(c, lim, d, lim);for (int i = 1; i <= m; i++) {pw[i] = quick_pow(a[i], mod - 2);}int ans = 0;for (int d = -1; d <= k; d++) {int gg = 1ll * c[d + k + 2] * B[d + 1] % mod * ifac[d + 1] % mod;for (int i = 1; i <= m; i++) {gg = 1ll * gg * (1 - pw[i] + mod) % mod;pw[i] = 1ll * pw[i] * a[i] % mod;}ans = (ans + gg) % mod;}printf("%d\n", ans);}return 0;
}