Race to 1 Again LightOJ - 1038

最新推荐文章于 2020-02-07 11:05:28 发布

墨碎江南

最新推荐文章于 2020-02-07 11:05:28 发布

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分类专栏：期望

本文链接：https://blog.youkuaiyun.com/dsaghjkye/article/details/83023862

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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

Sample Output

Case 1: 0

Case 2: 2.00

Case 3: 3.0333333333

要求将一个数变成1，每次可以除去它的任意一个因子，问你次数的期望。

设 f [ i ] 为 i 的期望

那么显然 f [ 1 ] =1 , f [ i ] = ( ( k + f [ i ] ) / x ) + 1

k 代表他所有因子的期望之和（不包括本身）

x 代表他所有因子的个数（包括本身）

为啥最后要加一，因为之所以期望变成所有因子期望和 / x 是因为已经进行了一次操作，期望最少为1

#include<bits/stdc++.h>
using namespace std;
double f[100010];
int  main()
{
   memset(f,0,sizeof(f));
   for(int i=2;i<=100000;i++)
   {
       double x=0;
       for(int j=1;j*j<=i;j++)
       {
           if(i%j==0)
           {
            x++;
            f[i]+=f[j];
            if(i/j!=j)
            {
                x++;
                f[i]+=f[i/j];
            }
           }
       }
       f[i]=(f[i]+x)/(x-1.0);
   }
   int n,t;
   scanf("%d",&t);
   for(int oo=1;oo<=t;oo++)
   {
    scanf("%d",&n);
    printf("Case %d: %.6lf\n",oo,f[n]);
   }
}