EOJ Monthly 2019.11

$$\begin{gathered} \sum _{i=1}^n\sum _{a_1=1}^i\sum _{a_2=1}^i\sum _{a_3=1}^i\cdots \sum _{a_{k-1}}^i\sum _{a_k}^i[gcd(a_1,a_2,a_3,\dots ,a_{k-1},a_k,i)==1] \\ =\sum _{i=1}^n\sum _{d\mid i}\mu (d)\lfloor \frac{i}{d}{\rfloor }^k \\ =\sum _{d=1}^n\mu (d)\sum _{d\mid i}\lfloor \frac{i}{d}{\rfloor }^k \\ =\sum _{d=1}^n\mu (d)\sum _{t=1}^{\frac{n}{d}}{t}^k \end{gathered}$$

然后杜教筛筛出$\sum _{i=1}^n\mu (i)$的前缀和，用拉格朗日插值得到$\sum _{i=1}^n{i}^k$这个式子，再加上数论分块即可完美解决。

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f;
const double eps = 1e-7;const int N = 1e6 + 10, mod = 998244353;ll fac[N], pre[N], suc[N], inv[N], prime[N], sum[N], mu[N], n, k, cnt;bool st[N];ll quick_pow(ll a, int n) {ll ans = 1;while(n) {if(n & 1) ans = ans * a % mod;a = a * a % mod;n >>= 1;}return ans;
}void init() {sum[1] = mu[1] = 1;for(int i = 2; i < N; i++) {if(!st[i]) {prime[cnt++] = i;mu[i] = -1;sum[i] = quick_pow(i, k);}for(int j = 0; j < cnt && i * prime[j] < N; j++) {st[i * prime[j]] = 1;sum[i * prime[j]] = 1ll * sum[i] * sum[prime[j]] % mod;if(i % prime[j] == 0) break;mu[i * prime[j]] = -mu[i];}}fac[0] = inv[0] = 1;for(int i = 1; i < N; i++) {mu[i] = (mu[i] + mu[i - 1]) % mod;sum[i] = (sum[i] + sum[i - 1]) % mod;fac[i] = 1ll * fac[i - 1] * i % mod;}inv[N - 1] = quick_pow(fac[N - 1], mod - 2);for(int i = N - 2; i >= 1; i--) {inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;}
}ll solve(ll n) {ll ans = 0;pre[0] = suc[k + 3] = 1;for(int i = 1; i <= k + 2; i++) pre[i] = 1ll * pre[i - 1] * (n - i) % mod;for(int i = k + 2; i >= 1; i--) suc[i] = 1ll * suc[i + 1] * (n - i) % mod;for(int i = 1; i <= k + 2; i++) {ll a = 1ll * pre[i - 1] * suc[i + 1] % mod, b = 1ll * inv[i - 1] * inv[k + 2 - i] % mod;if((k + 2 - i) & 1) b *= -1;ans = ((ans + 1ll * sum[i] * a % mod * b % mod) % mod + mod) % mod;}return ans;
}unordered_map<ll, ll> ans_s;ll S(ll n) {if(n < N) return mu[n];if(ans_s.count(n)) return ans_s[n];ll ans = 1;for(ll l = 2, r; l <= n; l = r + 1) {r = n / (n / l);ans = ((ans - 1ll *(r - l + 1) * S(n / l)) % mod + mod) % mod;}return ans_s[n] = ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);scanf("%lld %lld", &n, &k);init();ll ans = 0;for(ll l = 1, r; l <= n; l = r + 1) {r = n / (n / l);ans = (ans + 1ll * ((S(r) - S(l - 1)) % mod + mod) % mod * solve(n / l) % mod) % mod;}printf("%lld\n", ans);return 0;
}