题目
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
大意
给出一个数，每次随机处一个它的因子，求他变成1的期望次数。
题解
令 f[I] 表示i在f[i]此后变成1。
[f[i] = sum_{j|i} (f[j]+1) / k, (k = sum_j [j|i])
]
代码
#include<bits/stdc++.h>
#define repeat(a,b,c,g) for (int a=b,abck=(g>=0?1:-1);abck*(a)<=abck*(c);a+=g)
using namespace std;
double f[110000];
int main()
{
f[1] = 0;
for (int i=2;i<=100000;i++)
{
double tot = 0;
int tp = -1;
for (int j=1;j<=sqrt(i);j++)
{
if (i % j == 0)
{
tp ++, tot += f[j] + 1;
if (j * j != i)
tp ++, tot += f[i/j] + 1;
}
}
f[i] = tot / tp;
}
int n;
cin >> n;
for (int i=1;i<=n;i++)
{
int tmp;
cin >> tmp;
printf("Case %d: %.7f
",i,f[tmp]);
}
}