[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
",iCase++,dp[n]);
    }
    return 0;
}