【莫比乌斯反演】E - Sum of gcd of Tuples (Hard)-ABC162

题目链接(https://atcoder.jp/contests/abc162/tasks/abc162_e)

Time Limit: 2 sec / Memory Limit: 1024 MB

Score : $500$ points

Problem Statement

Consider sequences$\{A_1,...,A_N\}$ of length $N$ consisting of integers between $1$ and $K$ (inclusive).

There are $K^N$ such sequences. Find the sum of $\gcd(A_1, ..., A_N)$ over all of them.

Since this sum can be enormous, print the value modulo$(10^9+7)$.Here $\gcd(A_1, ..., A_N)$denotes the greatest common divisor of $A_1, ..., A_N$.

Constraints

$2 \leq N \leq 10^5$

$1\leq K \leq 10^5$

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the sum of $\gcd(A_1, ..., A_N)$over all $K^N$ sequences, modulo $(10^9+7)$.

Sample Input 1

3 2

Sample Output 1

9

$\gcd(1,1,1)+\gcd(1,1,2)+\gcd(1,2,1)+\gcd(1,2,2)+\gcd(2,1,1)+\gcd(2,1,2)+\gcd(2,2,1)+\gcd(2,2,2)=1+1+1+1+1+1+1+2=9$

Thus, the answer is 9.

Sample Input 2

3 200

Sample Output 2

10813692

Sample Input 3

100000 100000

Sample Output 3

742202979

Be sure to print the sum modulo$(10^9+7)$.

题意

给一个长度为$N$的序列，每一位的值都取$1$到$k$之间，求所有序列，（单序列）所有数位最大公因数之和，模$1e9+7$

思路

（苦脸）初学，很难搞懂，照葫芦画瓢，emo老师讲课笔记++；

第一反应应该枚举最大公因数$d$,则$1≤d≤k$，接下来再看有多少组 $\gcd(A_1, ..., A_N) == 1$

所以初步可以写出

$ans = \sum_{d=1}^kd\sum_{1≤A_i≤k}[gcd(A_1,...,A_N) = d]$

设$f(d) = \sum_{1≤A_i≤k}[gcd(A_1,...,A_N) = d]$

$g(d) = \sum_{d|d'}f(d') = \sum_{1≤A_i≤k}[d|A_1][d|A_2]...[d|A_k]$

$d$如果是$A_i$的倍数的话，他就有$\lfloor n/d \rfloor$种取法

那么求和可得$g(d) = \sum_{1≤A_i≤k}\lfloor n/d \rfloor^N$

就可以知道$f(d) = \sum_{d|d'}μ(d'/d)g(d') = \sum_{d|d'}μ(d'/d)\lfloor k/d' \rfloor^N$

代回原来的式子可得

$ans = \sum_{d=1}^kd\sum_{d|d'}μ(d'/d)\lfloor k/d' \rfloor^N$

置换两个求和号

$ans = \sum_{d'=1}^k\lfloor k/d' \rfloor^N\sum_{d|d'}dμ(d'/d)$

又因为$\sum_{d|d'}dμ(d'/d) =φ(d')$（相当于是用充斥的方法去计算$φ(d')$的数量）

证明：（暂时咕，马上补）

所以$ans = \sum_{d' = 1} ^ k\lfloor k/d' \rfloor^Nφ(d')$

复杂度$O(klogN)$

AC代码

#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define llinf 0x3f3f3f3f3f3f3f3f
#define int long long
#define ull unsigned long long
#define PII pair<int,int>
#define endl '\n'
using namespace std;
typedef long long ll;
const int N = 1e5 + 100;
const int mod = 1e9 + 7;
const double pi = acos(-1.0);
int t, n, k;
int a[N],b[N];
int gcd(int a, int b) {return b ? gcd(b, a % b) : a;}
int quickpow(int a, int b) {//快速幂
	int res = 1;
	while (b > 0) {
		if (b & 1) res = res * a % mod;
		b >>= 1;
		a = a * a % mod;
	}
	return res;
}
void init() {//初始化
	for (int i = 0; i <= n; i++) {
		a[i] = 0;
		b[i] = 0;
	}
	return;
}
void solve() {
	init();
	int ans = 0;
	for (int i = 1; i <= k; i++) {//求┕k/d'┙^N
		a[i] = quickpow(k / i, n) % mod;
	}
	for (int i = k; i >= 1; i--) {//求φ(d')
		b[i] = a[i];
		for (int j = i << 1; j <= k; j += i) {
			b[i] -= b[j];
		}
		b[i] = (b[i] + mod) % mod;
	}
	for (int i = 1; i <= k;i ++) {//相乘累加
		ans += (b[i] * i) % mod;
	}
	cout << (ans + mod) % mod << endl;//取模输出
	return;
}
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	while (cin >> n >> k) {
		solve();
	}
	return 0;
}