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]={0},b[maxn],t;
void init()
{
    long long i,j;
    t=0;
    for(i=2;i<maxn;i++)
    {
        if(!a[i])
        {
            b[t++]=i;
            j=i*i;
            while(j<maxn)
            {
                a[j]=1;
                j+=i;
            }
        }
    }
}
double fun(int n)
{
    double sum=0;
    int i,x=0,y=0;
    for(i=0;b[i]<=n&&i<t;i++)
    {
        x++;
        if(n%b[i]==0)
        {
            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=1;i<=t;i++)
    {
        scanf("%d",&n);
        printf("Case %d: %lf
",i,fun(n));
    }
}