Race to 1 Again LightOJ

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 (≤ 10000), denoting the number of test cases.

Each case begins with an integer N (1 ≤ N ≤ 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
Sample Output
Case 1: 0
Case 2: 2.00
Case 3: 3.0333333333

大致题意：告诉你一个正整数N，每次等概率的选择它的一个约数，然后除掉这个约数，更新N的值，问当N值变成1的期望次数。

思路：假设N的约数a[i]有num个，从小到大编号为1到num，那么a[1]=1，a[num]=N。易知，当一个数除以它的一个约数所得到的数一定是它的另一个约数。假设dp[i]表示将i值变成1所需的期望次数。
那么 dp[N]=(dp[a[1]]+1)/num+(dp[a[2]]+1)/num+….+(dp[a[num]]+1)/num
化解下得 dp[N]=(dp[a[1]]+dp[a[2]]+….+dp[a[num-1]]+num)/（num-1）,dp[1]=0。

代码如下

#include<bits/stdc++.h>const int N=1e5+5;#define LL long long intdouble dp[N];void init() {    dp[1]=0;    for(int i=2;i<N;i++)    {        int cnt=2;        double sum=0;        for(LL j=2;j*j<=i;j++)        {            if(i%j==0)            {                sum+=dp[j];                if(j*j<i)                {                    sum+=dp[i/j];                    cnt++;                }                cnt++;            }        }        dp[i]=(sum+cnt)/(cnt-1);    }}int main(){    init();    int T;    scanf("%d",&T);    for(int cas=1;cas<=T;cas++)    {        int n;        scanf("%d",&n);        printf("Case %d: %.10lf\n",cas,dp[n]);      }       return 0;}