「Light OJ1038」Race to 1 Again【期望dp】

1038 - Race to 1 Again

Time Limit: 2 second(s) Memory Limit: 32 MB

Rimi learned a new thing about integers, which is - any positive integer greater than $1$ can be divided by its divisors. So, he is now playing with this property. He selects a number $N$. And he calls this $D$.

In each turn he randomly chooses a divisor of $D$ ($1$ to $D$). Then he divides $D$ by the number to obtain new $D$. He repeats this procedure until $D$ becomes $1$. What is the expected number of moves required for $N$ to become $1$.

Input

Input starts with an integer $T(\le 10000)$, denoting the number of test cases.

Each case begins with an integer $N(1\le N\le 10^5)$.

Output

For each case of input you have to print the case number and the expected value. Errors less than $10^{-6}$ will be ignored.

Sample Input

3
1
2
50

Case 1: 0
Case 2: 2.00
Case 3: 3.0333333333

题意

• 给你一个整数，每次操作可以选一个$n$的正因子并除以这个因子，求这个数第一次变成$1$的期望操作数

解法

• 计$dp[i]$表示$n=i$时对应的答案
• 我们知道，期望是试验中每次可能结果的概率乘以其结果的总和，对应于本题中，$$dp[i]=\sum ^{所有使i变成1的操作方案}选择此方案对应的概率\times 此方案的操作数$$
  • 首先说明$dp[2]=2$分以下情况讨论
    • 第一步选择2的因子$2$，那么$n$直接变成$1$，结束操作，这种方案对期望的贡献是$T_1=\frac{1}{2}\times 1$
    • 第一步选择因子$1$，那么如果后面继续选择$1$，$n$是保持$2$不变的，直到选择$2$，那么显然这种方案下对应的方案是无穷多种的，其贡献
      $$\begin{array}{rl}{T}_2& =\underset{n\to \infty }{\lim }{\left(\frac{1}{2}\right)}^1\times \frac{1}{2}\times 2+{\left(\frac{1}{2}\right)}^2\times \frac{1}{2}\times 3+...+{\left(\frac{1}{2}\right)}^n\times \frac{1}{2}\times (n+1)\\ \frac{T_2}{2}& =\underset{n\to \infty }{\lim }{\left(\frac{1}{2}\right)}^2\times \frac{1}{2}\times 2+{\left(\frac{1}{2}\right)}^3\times \frac{1}{2}\times 3+...+{\left(\frac{1}{2}\right)}^{n+1}\times \frac{1}{2}\times (n+1)\\ 相减，得& \\ \frac{T_2}{2}& =\underset{n\to \infty }{\lim }{\left(\frac{1}{2}\right)}^1\times \frac{1}{2}\times 2+{\left(\frac{1}{2}\right)}^2\times \frac{1}{2}+...+{\left(\frac{1}{2}\right)}^n\times \frac{1}{2}-{\left(\frac{1}{2}\right)}^{n+1}\times \frac{1}{2}\times (n+1)\\ & =\frac{1}{2}+\underset{n\to \infty }{\lim }{\left(\frac{1}{2}\right)}^2\times \frac{1}{2}+...+{\left(\frac{1}{2}\right)}^n\times \frac{1}{2}\\ & =\frac{1}{2}+\underset{n\to \infty }{\lim }\frac{1}{2}\times \frac{\frac{1}{4}\times (1-(\frac{1}{2}{)}^{n-1})}{1-\frac{1}{2}}\\ & =0.75\\ T_2& =1.5\end{array}$$
    • 所以$dp[2]=T_1+T_2=2$
  • 然后就可以在可以递推了
    • 定义$num[i]$表示$i$的因子个数，$fac[i][j]$表示$i$的第$j$个因子
      $$dp[i]=\sum _{j=1}^{num[i]}\frac{1}{num[i]}\times (dp[fac[i][j]]+1)$$
      需要注意由于计算$dp[i]$的时候需要用到$dp[i]$，而$dp[i]$未求出，可以把右边的包含因子$i$的项移到左边，然后再计算$dp[i]$，相当于解一个一元一次方程

附代码

#include<bits/stdc++.h>

using namespace std;
const int maxn=1e5+10;

double dp[maxn];

int main()
{
    dp[1]=0,dp[2]=2.0;
    for(int i=3;i<maxn;i++){
        int num=0;
        for(int j=1;j*j<=i;j++){
            if(i%j==0){
                num++;
                if(j*j!=i) num++;
            }
        }
        dp[i]=num*1.0/(num-1);
        for(int j=1;j*j<=i;j++){
            if(i%j==0){
                dp[i]+=dp[j]/(num-1);
                if(j!=1&&j*j!=i) dp[i]+=dp[i/j]/(num-1);
            }
        }
    }
    int t;scanf("%d",&t);
    for(int i=1,n;i<=t;i++){
        scanf("%d",&n);
        printf("Case %d: %.10lf\n",i,dp[n]);
    }
}