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
题目的意思是给你一个数字n，每一次你可以选择一个n的一个约数然后除以他得到n‘’，一直到这个数字变成1
有期望公式我们知道有E(X) = x1*p（x1）+ x2 *p（x2） + x3*p（x3）+...................................+ xn*p(xn) ；
但是这个期望公式对这个题目没有任何的意义，假设这个数字有num个因数（包括他本身），那么这个数字到1
的次数的期望是E（X）= （E(a[1]) + E(a[2]) + E(a[3])+ .................+E(a[num])）/(num-1)//这里有必要说明一下
约数1的贡献是零（概率是1/(num-1)），另外还有就是E(a[num]) = tmp,就是x有几个约数，因为从x到约数需要一步
num个约数的话就要num步；
#include <bits/stdc++.h>
using namespace std;
const int Max = 1e5+100;
const int inf = 0x3f3f3f3f;
typedef long long ll;
#define rep(i,s,n) for( int i=s;i<=n;i++)
#define per(i,n,s) for( int i=n;i>=s;i--)
int t,tmp;
double dp[Max];
void pri(){
dp[1]=0.0,dp[2]=2.0;
rep(i,3,Max-10){
tmp=0;
double sum=0;
rep(j,1,sqrt(i)){
if(i%j!=0) continue;
sum+=dp[j], tmp++;
if(i/j!=j) sum+=dp[i/j],tmp++;
}
sum+=tmp;//表示的是从i到这个约数要除一次每一个约数都要除以一次有tmp个约数就要tmp次；
dp[i]=sum/(tmp-1);//因为1没有任何的贡献；
}
}
int main(){
int t;
pri();
scanf("%d",&t);
int ca=1;
while(t--){
int n;
scanf("%d",&n);
printf("Case %d: %.10lf\n",ca++,dp[n]);
}
return 0;
}