C

C - Race to 1 Again

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

分析：

$因为 所有数最终都会到达1，所以用dp[i]，表示i到1的的操作期望$

$dp[i]=sum _{d|i}(dp[d]+1)*(1/tot)，tot代表i的因子的个数$

当$d=i$的时候，这时候将等式右边移项即可

#include<cstdio>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<set>
#include<cmath>
#include<vector>
#include<cstring>
#include<string>
#include<iostream>
#include<iomanip>
#define mset(a,b)   memset(a,b,sizeof(a))
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxn=1e5+10;
const int branch=26;
const int inf=0x7fffffff;
const ll mod=1e9+7;
double dp[maxn];//n到i的期望
int tot;
void init()
{
    dp[1]=0;
    for(int k=2; k<=100000; ++k)
    {
        tot=2;
        dp[k]=2.0;
        for(int i=2; i*i<=k; ++i)//求出小于k的所有因子
            if(k%i==0)
            {
                dp[k]+=dp[i]+1;
                tot++;
                if(i!=k/i)
                {
                    dp[k]+=dp[k/i]+1;
                    tot++;
                }
            }
        dp[k]/=tot-1;
    }
}
int main()
{
    int cas=0;
    int n,t;
    init();
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        printf("Case %d: %.7f
",++cas,dp[n]);
    }
    return 0;
}