EOJ Monthly 2019.11 E. 数学题（反演 + 杜教筛 + 拉格朗日插值）

EOJ Monthly 2019.11

∑ i = 1 n ∑ a 1 = 1 i ∑ a 2 = 1 i ∑ a 3 = 1 i ⋯ ∑ a k − 1 i ∑ a k i [ g c d ( a 1 , a 2 , a 3 , … , a k − 1 , a k , i ) = = 1 ] = ∑ i = 1 n ∑ d ∣ i μ ( d ) ⌊ i d ⌋ k = ∑ d = 1 n μ ( d ) ∑ d ∣ i ⌊ i d ⌋ k = ∑ d = 1 n μ ( d ) ∑ t = 1 n d t k \sum_{i = 1} ^{n} \sum_{a_1 = 1} ^{i} \sum_{a_2 = 1} ^{i} \sum_{a_3 = 1} ^{i} \dots \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\\ i=1∑n​a1​=1∑i​a2​=1∑i​a3​=1∑i​⋯ak−1​∑i​ak​∑i​[gcd(a1​,a2​,a3​,…,ak−1​,ak​,i)==1]=i=1∑n​d∣i∑​μ(d)⌊di​⌋k=d=1∑n​μ(d)d∣i∑​⌊di​⌋k=d=1∑n​μ(d)t=1∑dn​​tk

然后杜教筛筛出 ∑ i = 1 n μ ( i ) \sum\limits_{i = 1} ^{n} \mu(i) i=1∑n​μ(i)的前缀和，用拉格朗日插值得到 ∑ i = 1 n i k \sum\limits_{i = 1} ^{n} i ^ k i=1∑n​ik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;
}