2019 南京网络赛 E. K Sum（莫比乌斯反演 + 杜教筛）

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化简 $$f_n(k)=\sum _{l_1=1}^n\sum _{l_2=1}^n...\sum _{l_k=1}^n(gcd(l_1,l_2,...,l_k){)}^2$$$$=\sum _{d=1}^n\sum _{l_1=1}^n\sum _{l_2=1}^n...\sum _{l_k=1}^n{d}^2\ast [gcd(l_1,l_2,...,l_k)=d]$$$$=\sum _{d=1}^n\sum _{l_1=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{l_2=1}^{\lfloor \frac{n}{d}\rfloor }...\sum _{l_k=1}^{\lfloor \frac{n}{d}\rfloor }{d}^2\ast [gcd(l_1,l_2,...,l_k)=1]$$$$=\sum _{d=1}^n{d}^2\sum _{l_1=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{l_2=1}^{\lfloor \frac{n}{d}\rfloor }...\sum _{l_k=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{p|gcd(l_1,l_2,...,l_k)}\mu (p)$$$$=\sum _{d=1}^n{d}^2\sum _{p=1}^{\lfloor \frac{n}{d}\rfloor }\mu (p)\sum _{l_1=1}^{\lfloor \frac{n}{p\ast d}\rfloor }\sum _{l_2=1}^{\lfloor \frac{n}{p\ast d}\rfloor }...\sum _{l_k=1}^{\lfloor \frac{n}{p\ast d}\rfloor }$$$$=\sum _{T=1}^n\sum _{d|T}{d}^2\ast \mu (\frac{T}{d})\ast (\lfloor \frac{n}{T}\rfloor {)}^k$$

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化简答案式子
$$\sum _{i=2}^k{f}_n(i)=\sum _{i=2}^k\sum _{T=1}^n\sum _{d|T}{d}^2\ast \mu (\frac{T}{d})\ast (\lfloor \frac{n}{T}\rfloor {)}^i$$$$=\sum _{T=1}^n\sum _{d|T}{d}^2\ast \mu (\frac{T}{d})\ast \sum _{i=2}^k(\lfloor \frac{n}{T}\rfloor {)}^i$$
后一项是等比数列求和，可用欧拉定理降幂，特判公比为1的情况。
令 $g(T)={\sum }_{d|T}{d}^2\ast \mu (\frac{T}{d})$，需要预处理 $g$的前缀和。
首先很明显 $g(T)$是一个积性函数
需要预处理大概 $10^6$的前缀和
（不会线性筛也可以用$n\log n$的暴力筛，比如我）
因为 $g=\mu \ast I{d}^2$，$g\ast I=\mu \ast I\ast I{d}^2=I{d}^2$（狄力克雷卷积满足交换律，结合律）

复杂度：$O(10^6\log 10^6+n^{\frac{2}{3}}+\sqrt{n}\log (10^9))$，第一项取决于筛的方法

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代码：

#include
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
char s[maxn];
bool ispri[maxn];
int pri[maxn],sum[maxn],mu[maxn];
int n,k,t;
map<int,int> mp;
int inv6;
void sieve(int n) {
	ispri[0] = ispri[1] = true;
	pri[0] = 0;mu[1] = 1;
	for(int i = 1; i <= n; i++) {
		if(!ispri[i]) pri[++pri[0]] = i,mu[i] = -1;
		for(int j = 1; j <= pri[0] && i * pri[j] <= n; j++) {
			ispri[i * pri[j]] = true;
			if(i % pri[j] == 0) break;
			mu[i * pri[j]] = -mu[i];
		}
	}
	for(int i = 1; i <= n; i++)
		for(int j = 1; i * j <= n; j++)
			sum[i * j] = (sum[i * j] + 1ll * i * i % mod * mu[j] % mod + mod) % mod;
	for(int i = 1; i <= n; i++)
		sum[i] = (sum[i] + sum[i - 1]) % mod;
}
int fpow(int a,int b) {
	int r = 1;
	while(b) {
		if(b & 1) r = 1ll * r * a % mod;
		a = 1ll * a * a % mod;
		b >>= 1;
	}
	return r;
}
int cal(int x) {
	return 1ll * x * (x + 1) % mod * (2 * x + 1) % mod * inv6 % mod;
}
int getsum(int n) {
	if(n <= maxn - 10) return sum[n];
	if(mp[n]) return mp[n];
	int res = cal(n);
	int i,j;
	for(i = 2; i <= n; i = j + 1) {
		j = n / (n / i);
		res = (res - 1ll * getsum(n / i) * (j - i + 1) % mod + mod) % mod;
	}
	return mp[n] = res;
}
int main() {
	sieve(maxn - 10);
	inv6 = fpow(6,mod - 2);
	scanf("%d",&t);
	while(t--) {
		scanf("%d%s",&n,&s);
		int len = strlen(s);
		k = 0;
		int k2 = 0;
		for(int i = 0; i < len; i++) {
			k = (1ll * k * 10 % (mod - 1) + s[i] - '0') % (mod - 1);
			k2 = (1ll * k2 * 10 % mod + s[i] - '0') % mod;		//用来处理公比为 1的情况
		}
		int l,r;
		int ans = 0;
		for(l = 1; l <= n; l = r + 1) {
			r = n / (n / l);
			int p = n / l;
			if(p == 1) 
				ans = (ans + 1ll * (k2 - 1) * ((getsum(r) - getsum(l - 1) + mod) % mod) % mod + mod) % mod;
			else {
				int q = (1 - p + mod) % mod;
				int inv = fpow(q,mod - 2);
				int tmp = (1ll * p * ((1 - fpow(p,k) + mod) % mod) % mod * inv % mod - p + mod) % mod;
				ans = (ans + 1ll * tmp * ((getsum(r) - getsum(l - 1) + mod) % mod) % mod) % mod;
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}