LIghtOJ1038---Race to 1 Again(概率dp)

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;
}