CF1139D Steps to One （莫比乌斯反演 期望dp）

\[f[1] = 0 \]

\[f[i] = 1 + \frac{1}{m} \sum_{j = 1} ^ n f[gcd(i, j)] \ \ \ \ \ \ (i != 1) \]

• 然后发现后面这一块gcd的个数只可能是i的约数， 那么考虑枚举约数

\[f[i] = 1 + \frac{1}{m}\sum_{d | i} f[d] cnt(d, i) \]

• \(cnt(d, i)\)表示和[1,m]内与i的gcd为d的数字个数
• 考虑这个东西能够怎么算， \(cnt(d, i)\)显然 等于 \(1\ \ to \ \ (m / d)\) 中 和（i / d)互质的数的个数， 后者是莫比乌斯反演的经典形式
• 然后暴力就能过了

/*



*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define ll long long 
#define M 100010
#define mmp make_pair
using namespace std;
const int mod = 1000000007;

void add(int &x, int y)
{
	x += y;
	x -= x >= mod ? mod : 0;
	x += x < 0 ? mod : 0; 
}

int mul(int a, int b)
{
	return 1ll * a * b % mod;
}

int poww(int a, int b)
{
	int ans = 1, tmp = a;
	for(; b; b >>= 1, tmp = mul(tmp, tmp)) if(b & 1) ans = mul(ans, tmp);
	return ans;
}

vector<int> to[M];
int f[M], mu[M], n, ans;

int read()
{
	int nm = 0, f = 1;
	char c = getchar();
	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
	return nm * f;
}

int main()
{
	n = read();
	mu[1] = 1;
	for(int i = 1; i <= n; i++)
	{
		for(int j = i; j <= n; j += i) 
		{	
			to[j].push_back(i);
			if(j != i) mu[j] -= mu[i];
		}
	}
	for(int i = 1; i <= n; i++)
	{
		int p = n / i;
		f[i] = mul(f[i] + p, poww(n - p, mod - 2));
		add(ans, f[i] + 1);
		for(int j = i + i; j <= n; j += i)
		{
			int d = j / i, s = 0;
			for(int k = 0; k < to[d].size(); k++)
			{
				int v = to[d][k];
				add(s, mul(mu[v], p / v));
			}
			add(f[j], mul(s, f[i] + 1));
		}
	}
	ans = mul(ans, poww(n, mod - 2));
	cout << ans << "\n";
	
	
	
	return 0;
}