題干：

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

題目大意：

給個數字Ｄ，我們可以選擇1~Ｄ中可以被Ｄ整除的因子，除以D得到一個新的D，再用新D除以它的因子得到又一個新D，按次操作除到D=1時結束，求除的次數的期望值。

p[D]表示從D除到1的期望次數，D的因子有cnt個（包括1和本身）

解題報告：

假設dp[n]代表從n開始退化到1的期望次數，p[n]代表n的因子的集合（vector），且已經排好序。

。

整理公式得

又因為（因為已經排好序了）

。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 1e5 + 5; int n,a[MAX]; double dp[MAX]; vector<int> vv[MAX]; void init() {for(int i = 1; i<MAX; i++) {for(int j = 1; j*j<=i; j++) {if((i%j) != 0) continue;vv[i].pb(j);if(j*j!=i) vv[i].pb(i/j);}} } void init2() {dp[1]=0;for(int i = 2; i<MAX; i++) {int up = vv[i].size();for(int j = 0; j<up; j++) {int v = vv[i][j];if(v == i) continue;dp[i] += dp[v]/(up-1);}dp[i]+=up*1.0/(up-1);} } int main() {init();init2();int t,iCase=0;cin>>t;while(t--) {scanf("%d",&n);printf("Case %d: %.8f\n",++iCase,dp[n]);}return 0 ; }

?

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