AtCoder Beginner Contest 162 E Sum of gcd of Tuples (Hard) 莫比乌斯反演做法

另一种容斥做法

反演做法(复习一下反演)

n个求和

$\sum_{i=1}^{k}\sum_{j=1}^{k}...\sum_{z=1}^{k} gcd(i,j...z)$

做个变换 $\sum_{x = 1}^{k}x\sum_{i=1}^{k}\sum_{j=1}^{k}...\sum_{z=1}^{k} [gcd(i,j...z) = x]$

设 $f(x) = \sum_{i=1}^{k}\sum_{j=1}^{k}...\sum_{z=1}^{k} [gcd(i,j...z) = x]$

设 $g(x) = \sum_{x | d}f(d) =\sum_{x | d}\sum_{i=1}^{k}\sum_{j=1}^{k}...\sum_{z=1}^{k} [gcd(i,j...z) = d]$

= $[(int)(\frac{k}{x})]^n$ gcd(i,j,…z)为x的倍数，每个位置取值 $(int)(\frac{k}{x})$个

反演一手 $f(x) = \sum_{x|d}{u(\frac{d}{x})}g(d) = \sum_{x|d}{u(\frac{d}{x})}[(int)(\frac{k}{x})]^n$

原式就 $= \sum_{x=1}^{k}x\sum_{x|d}{u(\frac{d}{x})}[(int)(\frac{k}{d})]^n$

令 $T = \frac{d}{x}$ , $d = T*x$

= $\sum_{x=1}^{k}x\sum_{T}{u(T)}[(int)(\frac{k}{T*x})]^n$

令 $D =x * T$

= $\sum_{D=1}^{k}[(int)(\frac{k}{T*x})]^n\sum_{T|D}{u(T)}*\frac{D}{T}$

= $\sum_{D=1}^{k}[(int)(\frac{k}{D})]^nphi(D)$

然后分块+预处理phi前缀和

#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
const int man = 2e5+10;
#define IOS ios::sync_with_stdio(0)
typedef long long ll;
const ll mod = 1e9+7;

bool vis[man];
int prime[man],phi[man];

void init(int maxn){
	phi[1] = 1;
	vis[1] = 1;
	int cnt = 0;
	for(int i = 2;i <= maxn;i++){
		if(!vis[i]){
			prime[cnt++] = i;
			phi[i] = i-1;
		}
		for(int j = 0;j < cnt && i * prime[j] <= maxn;j++){
			vis[i*prime[j]] = 1;
			phi[i*prime[j]] = phi[i]*phi[prime[j]];
			if(i%prime[j]==0){
				phi[i*prime[j]] = phi[i]*prime[j];
				break;
			}
		}
	}
	for(int i = 1;i <= maxn;i++){
		phi[i] = (phi[i-1] + phi[i])%mod;
	}
}

ll quick_mod(ll a,ll b){
	ll ans = 1;
	while(b){
		if(b&1)ans = ans * a % mod;
		a = a * a % mod;
		b >>= 1;
	}
	return ans;
}

int main() {
	#ifndef ONLINE_JUDGE
		//freopen("in.txt", "r", stdin);
		//freopen("out.txt","w",stdout);
	#endif
	int n,k;
	cin >> n >> k;
	init(k+5);
	ll ans = 0;
	for(int l = 1,r = 0;l <= k;l = r + 1){
		r = k / (k / l);
		ans = (ans + quick_mod(k/l,n)*(phi[r] - phi[l-1] + mod)%mod) % mod;
	}
	cout << ans << endl;
	return 0;
}