[LOJ 1038] Race to 1 Again

C - Race to 1 Again

Time Limit:2000MS      Memory Limit:32768KB      64bit IO Format:%lld & %llu

Description

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

设dp[i]表示i变成1的期望次数,则
dp[i]=(SUM(dp[j])/k)+1,j为i的因子,k为其因子个数
然而当取其因子为1时,j=i/1=i,所以：dp[i]=((SUM(dp[j'])+dp[i])/k)+1,j'为i除开因子i的因子
整理：dp[i]=(SUM(dp[j'])+k)/(k-1)
记忆化搜索即可。

#include <iostream>

#include <algorithm>

#include <cstring>

#include <cstdio>

using namespace std;

#define N 100000



double dp[N+10];



double dfs(int n)

{

    if(n==1) return dp[n]=0;

    if(dp[n]!=-1) return dp[n];

    int k=0;

    double s=0;

    for(int i=1;i*i<=n;i++)

    {

        if(n%i==0)

        {

            if(i*i!=n)

            {

                k+=2;

                if(i!=n) s+=dfs(i);

                if(n/i!=n) s+=dfs(n/i);

            }

            else

            {

                k+=1;

                if(i!=n) s+=dfs(i);

            }

        }

    }

    return dp[n]=(s+k)/(k-1);

}

int main()

{

    for(int i=0;i<=N;i++) dp[i]=-1;

    int T,iCase=1;

    int n;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d",&n);

        dfs(n);

        printf("Case %d: %.10f\n",iCase++,dp[n]);

    }

    return 0;

}