2019ICPC南京网络赛 E.K Sum 反演+杜教筛

E.K Sum

Problem

Def. $$f_n(k)=\sum _{l_1=1}^n\sum _{l_2=1}^n...\sum _{l_k=1}^n(gcd(l_1,l_2,...,l_k){)}^2$$
Give n, Calculate $$\sum _{i=2}^k{f}_n(i)(\mathrm{m}\mathrm{o}\mathrm{d}1e9+7)$$

$1\le n\le 10^9$
$2\le k\le 10^{10^5}$

Solution

$$\begin{array}{rl}{f}_n(k)& =\sum _{i=1}^n\sum _{l_1=1}^n\sum _{l_2=1}^n...\sum _{l_k=1}^n{i}^2\cdot [(l_1,l_2,...,l_k)=i]\\ & =\sum _{i=1}^n\sum _{l_1=1}^{\lfloor \frac{n}{i}\rfloor }\sum _{l_2=1}^{\lfloor \frac{n}{i}\rfloor }...\sum _{l_k=1}^{\lfloor \frac{n}{i}\rfloor }{i}^2\cdot [(l_1,l_2,...,l_k)=1]\end{array}$$
反演可得
$$\begin{array}{rl}{f}_n(k)& =\sum _{i=1}^n\sum _{l_1=1}^{\lfloor \frac{n}{i}\rfloor }\sum _{l_2=1}^{\lfloor \frac{n}{i}\rfloor }...\sum _{l_k=1}^{\lfloor \frac{n}{i}\rfloor }{i}^2\sum _{d|(l_1,l_2,...,l_k)}\mu (d)\\ & =\sum _{i=1}^n{i}^2\sum _{d=1}^{\lfloor \frac{n}{i}\rfloor }\mu (d)\cdot \sum _{l_1=1}^{\lfloor \frac{n}{i\cdot d}\rfloor }\sum _{l_2=1}^{\lfloor \frac{n}{i\cdot d}\rfloor }...\sum _{l_k=1}^{\lfloor \frac{n}{i\cdot d}\rfloor }1\\ & =\sum _{i=1}^n{i}^2\sum _{d=1}^{\lfloor \frac{n}{i}\rfloor }\mu (d)\cdot \lfloor \frac{n}{i\cdot d}{\rfloor }^k\\ & =\sum _{l=1}^n\lfloor \frac{n}{l}{\rfloor }^k\sum _{d|l}\mu (d)\cdot (\frac{l}{d}{)}^2\end{array}$$
对$f_n(i)$求前缀和得
$$\begin{array}{rl}\sum _{i=2}^k{f}_n(i)& =\sum _{i=2}^k\sum _{l=1}^n\lfloor \frac{n}{l}{\rfloor }^i\sum _{d|l}\mu (d)\cdot (\frac{l}{d}{)}^2\\ & =\sum _{l=1}^n(\sum _{i=2}^k\lfloor \frac{n}{l}{\rfloor }^i)(\sum _{d|l}\mu (d)\cdot (\frac{l}{d}{)}^2)\\ & =\sum _{l=1}^n{T}_1\cdot T_2\end{array}$$
对于$T_1$，先用欧拉定理对k降幂后，可以使用等比数列求和公式直接求得，并特殊处理公比为1得情况。
对于$T_2$，用杜教筛。令$$g(n)=\sum _{d|n}\mu (d)\cdot (\frac{n}{d}{)}^2$$
由于$\mu ,id$都是积性函数，所以它们的狄利克雷卷积$g(n)$为积性函数。
因此
$$\begin{array}{rl}g(1)& =1\\ g(p)& =p^2-1\\ g(p^{k+1})& =p^{2k+2}-p^{2k}=g(p)\cdot p^{2k}\end{array}$$
则以上部分可用欧拉筛求得5e6内的值。
由于$$g\times I=\mu \times i{d}^2\times I=\mu \times I\times i{d}^2=i{d}^2$$
则对其求前缀和有
$$\frac{n(n+1)(2n+1)}{6}=\sum _{i=1}^n{i}^2=\sum _{i=1}^n\sum _{d|i}\mu (d)\cdot (\frac{i}{d}{)}^2=\sum _{\frac{i}{d}=1}^n\sum _{d=1}^{\lfloor \frac{n}{\frac{i}{d}}\rfloor }\mu (d)\cdot (\frac{i}{d}{)}^2=\sum _{i=1}^{n}G(\lfloor \frac{n}{i}\rfloor )$$
即$$G(n)=\frac{n(n+1)(2n+1)}{6}-\sum _{i=2}^{n}G(\lfloor \frac{n}{i}\rfloor )$$
总体复杂度$O(n^{\frac{2}{3}}+\sqrt{n}\cdot log10^9)$

// Cease to struggle and you cease to live
// E.cpp
// Created by Nickwzk on 2019_09_22 13:20
#include <bits/stdc++.h>
using namespace std;

typedef long long LL;
const int MAXN = 5e6 + 5, mod = 1e9 + 7;
int prime[MAXN], cnt;
LL g[MAXN], G[MAXN];
int fac[MAXN];
LL fac_pow[MAXN];
bool isp[MAXN];
unordered_map<LL, int> mp;
void Euler(int n) {
	fac_pow[1] = g[1] = fac[1] = 1;
	for(int i = 2; i <= n; i++) {
		if(!isp[i]) {
			prime[++cnt] = i;
			fac_pow[i] = fac[i] = i;
			g[i] = ((LL)i * i - 1) % mod;
		}
		for(int j = 1; j <= cnt && i * prime[j] <= n; j++) {
			isp[i * prime[j]] = 1;
			if(i % prime[j] == 0) {
				fac[i * prime[j]] = fac[i];
				fac_pow[i * prime[j]] = fac_pow[i] * prime[j];
				g[i * prime[j]] = g[fac[i]] * g[i / fac_pow[i]] % mod * fac_pow[i] % mod * fac_pow[i] % mod;
				break;
			}
			fac_pow[i * prime[j]] = fac[i * prime[j]] = prime[j];
			g[i * prime[j]] = g[i] * g[prime[j]] % mod;
		}
	}
	for(int i = 1; i <= n; i++) {
		G[i] = (G[i - 1] + g[i]) % mod;
	}
}
LL qpow(LL x, LL y) {
	LL res = 1;
	while(y) {
		if(y & 1) res = res * x % mod;
		x = x * x % mod;
		y >>= 1;
	}
	return res;
}
LL inv(LL x) {
	x = (x + mod) % mod;
	return qpow(x, mod - 2);
}
LL cal(LL n) {
	if(n < MAXN) return G[n];
	if(mp.count(n)) return mp[n];
	LL res = 0;
	for(LL i = 2, last; i <= n; i = last + 1) {
		last = n / (n / i);
		res += (last - i + 1) % mod * cal(n / i) % mod;
		res %= mod;
	}
	return mp[n] = ((__int128)n * (n + 1) * (2 * n + 1) / 6 % mod + mod - res) % mod;
}
inline LL get(LL n, LL x, LL y) {
	if(n == 1) return y;
	return n * n % mod * (mod + 1 - qpow(n, x)) % mod * inv(1 - n) % mod;
}
int main() {
	ios::sync_with_stdio(0); cin.tie(0); cout.precision(6); cout << fixed;
	Euler(MAXN - 1);
	int T; cin >> T;
	LL n; string k;
	LL x, y;
	while(T--) {
		cin >> n >> k;
		x = y = 0;
		for(auto c : k) {
			x = x * 10 + c - '0';
			y = y * 10 + c - '0';
			x %= mod - 1;
			y %= mod;
		}
		x = (x - 1 + mod - 1) % (mod - 1);
		y = (y - 1 + mod) % mod;
		LL ans = 0;
		for(LL i = 1, last; i <= n; i = last + 1) {
			last = n / (n / i);
			ans += get(n / i % mod, x, y) * (cal(last) - cal(i - 1) + mod) % mod;
			ans %= mod;
		}
		cout << ans << endl;
	}
	return 0;
}