【LightOJ - 1038】Race to 1 Again（概率dp，数学期望）

题干：

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 ≤ 105).

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

题目大意：

给个数字Ｄ，我们可以选择1~Ｄ中可以被Ｄ整除的因子，除以D得到一个新的D，再用新D除以它的因子得到又一个新D，按次操作除到D=1时结束，求除的次数的期望值。

p[D]表示从D除到1的期望次数，D的因子有cnt个（包括1和本身）

解题报告：

假设dp[n]代表从n开始退化到1的期望次数，p[n]代表n的因子的集合（vector），且已经排好序。

。

整理公式得

又因为（因为已经排好序了）

。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 1e5 + 5;
int n,a[MAX];
double dp[MAX];
vector<int> vv[MAX];
void init() {
  for(int i = 1; i<MAX; i++) {
    for(int j = 1; j*j<=i; j++) {
      if((i%j) != 0) continue;
      vv[i].pb(j);
      if(j*j!=i) vv[i].pb(i/j);
    }
  }
}
void init2() {
  dp[1]=0;
  for(int i = 2; i<MAX; i++) {
    int up = vv[i].size();
    for(int j = 0; j<up; j++) {
      int v = vv[i][j];
      if(v == i) continue;
      dp[i] += dp[v]/(up-1);
    }
    dp[i]+=up*1.0/(up-1);
  }
}
int main()
{
  init();
  init2();
  int t,iCase=0;
  cin>>t;
  while(t--) {
    scanf("%d",&n);
    printf("Case %d: %.8f\n",++iCase,dp[n]);
  }
  return 0 ;
}