【Codeforces1139D_CF1139D】Steps to One (Mobius_DP)

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(xin[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 (lfloorfrac{m}{n} floor), so we have:

[(m-lfloorfrac{m}{n} floor)f[n]=m+sum_{d|n,d eq n}f[d]cdot g(n, d) ]

Now the problem become how to calculate (g(n,d)). According to the defination,

[egin{aligned} g(n, d)&=sum_{i=1}^m[gcd(n, i)=d]\ &=sum_{i=1}^{lfloorfrac{m}{d} floor}[gcd(frac{n}{d},i)=1]\ &=sum_{i=1}^{lfloorfrac{m}{d} floor}epsilonleft(gcd(frac{n}{d},i) ight)\ end{aligned} ]

where (epsilon(x)=egin{cases}1 (x=1)\0 mathrm{otherwise}end{cases}) .

According to the Mobius Theorem ( (mu * 1 = epsilon) ) :

[egin{aligned} g(n,d)&=sum_{i=1}^{lfloorfrac{m}{d} floor}sum_{t|frac{n}{d},t|i}mu(t)\ &=sum_{t|frac{n}{d}}mu(t)cdot lfloor frac{m}{dt} floor end{aligned} ]

Let's return to (f[n]):

[(m-lfloorfrac{m}{n} floor)f[n]=m+sum_{d|n,d eq n}f[d]sum_{t|frac{n}{d}}mu(t)cdot lfloor frac{m}{dt} floor ]

Preprocess the divisors of all integer (x(xin[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 imes 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();
}