LightOJ 1038 Race to 1 Again

LightOJ 1038 Race to 1 Again

题目链接

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 ≤ 1e5).

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

典型的概率 DP ~
我们设 $dp[n]$ 为 $n->1$ 的步数期望，设 $n$ 有 $m$ 个因子：$f_1,f_2,\cdots ,f_m$，则有状态转移方程：
$dp[n]=dp[f_1]+1$
$dp[n]=dp[f_2]+1$
$\cdots$
$dp[n]=dp[f_m]+1$
累加得：
$m\ast dp[n]=dp[f_1]+dp[f_2]+\cdots +dp[f_m]+m$
我们知道 $f_m=n，f_1=1$，所以化简得：
$(m-1)dp[n]=dp[f_2]+dp[f_3]+\cdots +dp[f_{m-1}]+m$
最终得出状态转移方程：
$$dp[n]=\frac{m\ast (dp[f_2]+dp[f_3]+\cdots +dp[f_{m-1}])}{m-1}$$
因为 $n$ 不大，我们可以预处理出所有答案，AC代码如下：

#include
using namespace std;
typedef long long ll;
const int N=1e5+5;
double dp[N];
void pre(){
    dp[1]=0;
    for(int i=2;i<=N;i++){
        int num=0;
        for(int j=1;j<=sqrt(i);j++){
            if(i%j==0){
                dp[i]+=dp[j];
                num++;
                if(i/j!=j){
                    num++;
                    dp[i]+=dp[i/j];
                }
            }
        }
        dp[i]=(dp[i]+num)/(num-1);
    }
}
int main(){
    pre();
    int n,t;
    scanf("%d",&t);
    for(int i=1;i<=t;i++){
        scanf("%d",&n);
        printf("Case %d: %.6f\n",i,dp[n]);
    }
	return 0;
}