Problem:
Analysis:
After ACing E, I gave up D and spent the left 30 minutes chatting with Little Dino.
Let \(f[n]\) be the expected number of steps needed to make the greatest common divisor (gcd) become \(1\) when the gcd is \(n\) now, and \(g(n,d)\) be the number of \(x(x\in[1,m])\) that \(gcd(x, n)=d\) . So we have:
\[f[n]=1+\sum_{d|n}\frac{f[d]\cdot g(n, d)}{m}\]
To make it easy, multiply \(m\) to the equation:
\[mf[n]=m+\sum_{d|n}f[d]\cdot g(n, d)\]
Notice that \(d\) can be \(n\), and \(g(n,n)\) is \(\lfloor\frac{m}{n}\rfloor\), so we have:
\[(m-\lfloor\frac{m}{n}\rfloor)f[n]=m+\sum_{d|n,d\neq n}f[d]\cdot g(n, d)\]
Now the problem become how to calculate \(g(n,d)\). According to the defination,
\[ \begin{aligned} g(n, d)&=\sum_{i=1}^m[gcd(n, i)=d]\\ &=\sum_{i=1}^{\lfloor\frac{m}{d}\rfloor}[gcd(\frac{n}{d},i)=1]\\ &=\sum_{i=1}^{\lfloor\frac{m}{d}\rfloor}\epsilon\left(gcd(\frac{n}{d},i)\right)\\ \end{aligned} \]
where \(\epsilon(x)=\begin{cases}1\ (x=1)\\0\ \mathrm{otherwise}\end{cases}\) .
According to the Mobius Theorem ( \(\mu * 1 = \epsilon\) ) :
\[ \begin{aligned} g(n,d)&=\sum_{i=1}^{\lfloor\frac{m}{d}\rfloor}\sum_{t|\frac{n}{d},t|i}\mu(t)\\ &=\sum_{t|\frac{n}{d}}\mu(t)\cdot \lfloor \frac{m}{dt} \rfloor \end{aligned} \]
Let's return to \(f[n]\):
\[(m-\lfloor\frac{m}{n}\rfloor)f[n]=m+\sum_{d|n,d\neq n}f[d]\sum_{t|\frac{n}{d}}\mu(t)\cdot \lfloor \frac{m}{dt} \rfloor\]
Preprocess the divisors of all integer \(x(x\in[1,m])\) and then calculate \(f[n]\) as the equation above directly. Because the number of divisors of most integers is very small ( for integers not more than \(100000\), the maximum is \(128\) and the total number is about \(10^6\) to \(2\times 10^6\)) , so it won't TLE.
At last, the answer is:
\[ans=1+\sum_{i=1}^{m}\frac{f[i]}{m}\]
Code:
#include <cstdio>
#include <cstring>
#include <cctype>
#include <algorithm>
#include <vector>
using namespace std;
namespace zyt
{
typedef long long ll;
const int N = 1e5 + 10, p = 1e9 + 7;
vector<int> fac[N];
int n, f[N], pcnt, prime[N], mu[N];
bool mark[N];
void init()
{
for (int i = 1; i <= n; i++)
for (int j = 1; j * j <= i; j++)
if (i % j == 0)
{
fac[i].push_back(j);
if (j * j != i)
fac[i].push_back(i / j);
}
mu[1] = 1;
for (int i = 2; i <= n; i++)
{
if (!mark[i])
prime[pcnt++] = i, mu[i] = p - 1;
for (int j = 0; j < pcnt && (ll)i * prime[j] <= n; j++)
{
int k = i * prime[j];
mark[k] = true;
if (i % prime[j] == 0)
{
mu[k] = 0;
break;
}
else
mu[k] = p - mu[i];
}
}
}
int power(int a, int b)
{
int ans = 1;
while (b)
{
if (b & 1)
ans = (ll)ans * a % p;
a = (ll)a * a % p;
b >>= 1;
}
return ans;
}
int inv(const int a)
{
return power(a, p - 2);
}
int work()
{
scanf("%d", &n);
init();
f[1] = 0;
int ans = 0;
for (int i = 2; i <= n; i++)
{
for (int j = 0; j < fac[i].size(); j++)
{
int d = fac[i][j];
if (d == i)
continue;
int tmp = 0;
for (int k = 0, size = fac[i / d].size(); k < size; k++)
{
int t = fac[i / d][k];
tmp = (tmp + (ll)mu[t] * (n / d / t) % p) % p;
}
f[i] = (f[i] + (ll)tmp * f[d] % p) % p;
}
f[i] = (ll)(f[i] + n) * inv(n - n / i) % p;
}
for (int i = 1; i <= n; i++)
ans = (ans + f[i]) % p;
printf("%d", int(((ll)ans * inv(n) % p) + 1) % p);
return 0;
}
}
int main()
{
return zyt::work();
}