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Problem:

Codeforces 1139D

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();
}

【Codeforces1139D_CF1139D】Steps to One (Mobius_DP)

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原文地址：https://www.cnblogs.com/zyt1253679098/p/10584706.html