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
Output for Sample Input
3
1
2
50
Case 1: 0
Case 2: 2.00
Case 3: 3.0333333333
Problem Setter: Jane Alam Jan
dp[i]表示把i变成1的期望次数
/*************************************************************************
> File Name: c.cpp
> Author: ALex
> Mail: [email protected]
> Created Time: 2015年04月29日 星期三 19时40分52秒
************************************************************************/
#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>
using namespace std;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
double dp[100110];
double dfs(int num) {
if (dp[num] != -1) {
return dp[num];
}
int cnt = 2;
double ans = 0;
for (int i = 2; i * i <= num; ++i) {
if (num % i == 0) {
++cnt;
ans += dfs(num / i);
if (num / i != i) {
ans += dfs(i);
++cnt;
}
}
}
ans += cnt;
ans /= (cnt - 1);
return dp[num] = ans;
}
int main() {
int t;
scanf("%d",&t);
int icase = 1;
while (t--) {
int n;
scanf("%d", &n);
for (int i = 1; i <= n; ++i) {
dp[i] = -1;
}
dp[1] = 0;
printf("Case %d: %.12f\n", icase++, dfs(n));
}
return 0;
}