[概率DP] C - Race to 1 Again LightOJ

本文链接：https://blog.youkuaiyun.com/ummmmm/article/details/84545711

本文探讨了一个数学游戏，目标是找到将任意正整数通过不断除以其因数直至变为1所需的平均步骤数。游戏从选择一个数字N开始，随后随机选取其因数进行除法操作，直到N减小到1。文章提供了算法实现，采用动态规划方法预先计算每个数字的期望步数。

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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

Sample Output

Case 1: 0

Case 2: 2.00

Case 3: 3.0333333333

dp[6] = (dp[6] + dp[3] + dp[2] + dp[1]) / 4 + 1, 每次都要多一次

#include <bits/stdc++.h>
using namespace std;
const int mn = 1e5 + 10;

double dp[mn];
int main()
{
	dp[1] = 0;
	for (int i = 2; i < mn; i++)
	{
		double sum = 0;
		int j = 1, cnt = 0;
		for (j = 1; j * j <= i; j++)
		{
			if (i % j == 0 && j * j != i)
			{
				cnt += 2;
				sum = sum + dp[j] + dp[i / j];
			}
			else if (i % j == 0 && j * j == i)
				cnt++, sum += dp[j];
		}
		dp[i] = (sum + cnt) / (cnt - 1);
	}
	
	int T;
	scanf("%d", &T);
	for (int cas = 1; cas <= T; cas++)
	{
		int n;
		scanf("%d", &n);
		printf("Case %d: %.8f\n", cas, dp[n]);
	}
	return 0;
}