LightOJ 1038 Race To 1 Again(概率DP)

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

题解：题意就是给你一个数，然后你每次可以执行的操作为，将这个数除以它的因子（包含1和他本身），然后再将结果赋给他本身，直到变成1；

让你求执行次数的期望；

设DP[num]:表示数字num变为1执行次数的期望；

则： dp[num]=(dp[x1]+1)/n+(dp[x2]+1)/n+(dp[x3]+1)/n+...+(dp[xn]+1)/n;(其中xn==num)

则dp[num]=(sigma(1,n-1) xi + n)/(n-1);

参考代码：

#include<bits/stdc++.h>
using namespace std;
#define clr(a,val) memset(a,val,sizeof(a))
const int maxn=1e5+10;
int T,n;
double dp[maxn];

void prework()
{
    clr(dp,0);
    for(int i=2;i<maxn;++i)
    {
        int step=sqrt(i),num=-1;
        double sum=0;
        for(int j=1;j<=step;++j)
        {
            if(i%j==0)
            {
                sum+=dp[j]+1,num++;
                if(j!=i/j) sum+=dp[i/j]+1,num++;    
            } 
        }
        dp[i]=sum*1.0/num;
    }
}

int main()
{
    prework();
    scanf("%d",&T);
    for(int cas=1;cas<=T;++cas)
    {
        scanf("%d",&n);
        printf("Case %d: %.7lf
",cas,dp[n]);
    }
    
    return 0;
}