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$T=\frac{d}{x}$ ,$d=T\ast x$

= ${\sum }_{x=1}^{k}x{\sum }_{T}u(T)[(int)(\frac{k}{T\ast x}){]}^n$

令$D=x\ast T$

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

= ${\sum }_{D=1}^k[(int)(\frac{k}{D}){]}^{n}phi(D)$

然后分块+预处理phi前缀和

#pragma GCC optimize(2)
#include
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;
}