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
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。
令 f[I] 表示i在f[i]此后变成1。
\[ f[i] = \sum_{j|i} (f[j]+1) / k, (k = \sum_j [j|i]) \]
#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\n",i,f[tmp]); } }