「题解」Codeforces 1139D Steps to One

D. Steps to One

Description

• 给一个数列，每次随机选一个 \(1\) 到 \(m\) 之间的数加在数列末尾，数列中所有数的 \(\gcd = 1\) 时停止，求期望长度 \(\bmod 10^9 + 7\)。
• \(1\le m\le 100000\)。

Solution

设 \(E(x)\) 为 \(x\) 的期望值，\(P(x)\) 为事件 \(x\) 发生的概率。

则

\[\begin{aligned} E(len) & = \sum_{i \ge 1} P(len = i) \cdot i \\ & = \sum_{i \ge 1} P(len = i) \sum_{j = 1}^i 1 \\ & = \sum_{j \ge 1} \sum_{i\ge j} P(len = i) \\ & = \sum_{i \ge 1} P(len \ge i) \end{aligned} \]

发现

\[\sum_{i\ge 1} P(len = i) \]

恰好就是整体情况，为 \(1\)。

所以

\[\begin{aligned} E(len) & = \sum_{i \ge 1} P(len \ge i) \\ & = 1 + \sum_{i \ge 1} P(len > i) \end{aligned} \]

要使得 \(len > i\)，则必须满足前 \(i\) 个数的 \(\gcd > 1\)。

于是

\[P(len > i) = P\left(\gcd_{j = 1}^i\{a_j\} > 1 \right) \]

进行一个反面考虑

\[P\left(\gcd_{j = 1}^i \{a_j\} > 1 \right) = 1 - P\left(\gcd_{j = 1}^i \{a_j\} = 1 \right) \]

那么

\[\begin{aligned} P(len > i) & = 1 - P\left(\gcd_{j = 1}^i \{a_j\} = 1 \right) \\ & = 1 - \dfrac{\sum_{a_1 = 1}^m \sum_{a_2 = 1}^m \cdots \sum_{a_i = 1}^m [\gcd_{j = 1}^i \{ a_j\} = 1]}{m^i} \\ & = 1 - \dfrac{\sum_{d = 1}^m \mu(d) \left\lfloor\dfrac{m}{d}\right\rfloor^i}{m^i} \end{aligned} \]

单独将 \(d = 1\) 的情况拎出来，其值为 \(1\)。

\[P(len > i) = - \dfrac{\sum_{d = 2}^m \mu(d) \left\lfloor\dfrac{m}{d}\right\rfloor^i}{m^i} \]

代回原式

\[\begin{aligned} E(len) & = 1 + \sum_{i \ge 1} P(i > len) \\ & = 1 + \sum_{i \ge 1} - \dfrac{\sum_{d = 2}^m \mu(d) \left\lfloor\dfrac{m}{d}\right\rfloor^i}{m^i} \\ & = 1 - \sum_{i \ge 1} \dfrac{1}{m^i} \sum_{d = 2}^m \mu(d) \left\lfloor\dfrac{m}{d}\right\rfloor^i \\ & = 1 - \sum_{d = 2}^m \mu(d) \sum_{i \ge 1} \left(\dfrac{\left\lfloor\frac{m}{d}\right\rfloor}{m} \right)^i \end{aligned} \]

后面有一个无穷项等比数列求和

\[S = x + x^2 + x^3 + \cdots \\ xS = x^2 + x^3 + x^4 + \cdots \\ S - xS = x \\ S = \dfrac{x}{1 - x} \]

所以

\[\begin{aligned} E(len) & = 1 - \sum_{d = 2}^m \mu(d) \dfrac{\frac{\left\lfloor\frac{m}{d}\right\rfloor}{m}}{1 - \frac{\left\lfloor\frac{m}{d}\right\rfloor}{m}} \\ & = 1 - \sum_{d = 2}^m \mu(d) \dfrac{\left\lfloor\frac{m}{d}\right\rfloor}{m - \left\lfloor\frac{m}{d}\right\rfloor} \end{aligned} \]

\(\Theta(m)\) 计算即可，注意 \(m\) 比较小所以不需要整除分块。

Code

// 18 = 9 + 9 = 18.
#include <iostream>
#include <cstdio>
#include <cstring>
#define Debug(x) cout << #x << "=" << x << endl
typedef long long ll;
using namespace std;

namespace IO
{
	int len = 0;
	char buf[(1 << 20) + 1], *S, *T;
	#if ONLINE_JUDGE
		#define Getchar() (S == T ? T = (S = buf) + fread(buf, 1, (1 << 20) + 1, stdin), (S == T ? EOF : *S++) : *S++)
	#else
		#define Getchar() getchar()
	#endif
	#define re register
	inline int read()
	{
		re char c = Getchar();
		re int x = 0;
		while (c < '0' || c > '9')
			c = Getchar();
		while (c >= '0' && c <= '9')
			x = (x << 3) + (x << 1) + (c ^ 48), c = Getchar();
		return x;
    }
}
using IO::read;

const int MAXN = 1e5 + 5;
const int MOD = 1e9 + 7;
int add(int a, int b) {return (a + b) % MOD;}
int sub(int a, int b) {return (a - b + MOD) % MOD;}
int mul(int a, int b) {return (ll)a * b % MOD;}

int p[MAXN], mu[MAXN], inv[MAXN];
bool vis[MAXN];

void pre(int n)
{
	mu[1] = 1;
	for (int i = 2; i <= n; i++)
	{
		if (!vis[i])
		{
			p[++p[0]] = i;
			mu[i] = -1;
		}
		for (int j = 1; j <= p[0] && i * p[j] <= n; j++)
		{
			vis[i * p[j]] = true;
			if (i % p[j] == 0)
			{
				mu[i * p[j]] = 0;
				break;
			}
			mu[i * p[j]] = mu[i] * mu[p[j]];
		}
	}
	inv[1] = 1;
	for (int i = 2; i <= n; i++)
	{
		inv[i] = mul(MOD - MOD / i, inv[MOD % i]);
	}
}

int main()
{
	int m = read();
	pre(m);
	int res = 0;
	for (int d = 2; d <= m; d++)
	{
		res = add(res, mu[d] * mul(m / d, inv[m - m / d]));
	}
	printf("%d\n", sub(1, res));
	return 0;
}