Race to 1 概率dp

Race to 1

Time Limit: 10000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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Description

 

B

Race to 1 Input: Standard Input Output: Standard Output

Dilu have learned a new thing about integers, which is - any positive integer greater than 1 can be divided by at least one prime number less than or equal to that number. 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 prime number less than or equal to D. If D is divisible by the prime number then he divides D by the prime number to obtain new D. Otherwise he keeps the old D. He repeats this procedure until D becomes 1. What is the expected number of moves required for N to become 1.

[We say that an integer is said to be prime if its divisible by exactly two different integers. So, 1 is not a prime, by definition. List of first few primes are 2, 3, 5, 7, 11, …]

Input

Input will start with an integer T (T <= 1000), which indicates the number of test cases. Each of the next T lines will contain one integer N (1 <= N <= 1000000).

Output

For each test case output a single line giving the case number followed by the expected number of turn required. Errors up to 1e-6 will be accepted.

Sample Input                             Output for Sample Input

3 1 3 13

Case 1: 0.0000000000 Case 2: 2.0000000000 Case 3: 6.0000000000

 #include <algorithm>
 #include <cstdio>
 #include <iostream>
 #include <string.h>
 #include <math.h>
 #include <vector>
 using namespace std;
 #define maxn 1000010
 int a[maxn]={},b[maxn],t;
 void init()
 {
     long long i,j;
     t=;
     for(i=;i<maxn;i++)
     {
         if(!a[i])
         {
             b[t++]=i;
             j=i*i;
             while(j<maxn)
             {
                 a[j]=;
                 j+=i;
             }
         }
     }
 }
 double fun(int n)
 {
     double sum=;
     int i,x=,y=;
     for(i=;b[i]<=n&&i<t;i++)
     {
         x++;
         if(n%b[i]==)
         {
             y++;
             sum+=fun(n/b[i]);
         }
     }
     sum+=x;
     if(y)
     return sum/y;
     else return sum;
 }
 int main()
 {
     init();
     int i,j,t,n;
     scanf("%d",&t);
     for(i=;i<=t;i++)
     {
         scanf("%d",&n);
         printf("Case %d: %lf\n",i,fun(n));
     }
 }