「题解」Codeforces 1139D Steps to One
作者：互联网
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;
}
标签：right,题解,sum,len,ge,1139D,Steps,dfrac,left 来源： https://www.cnblogs.com/mangoworld/p/Solution-Codeforces-1139D.html