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最新推荐文章于 2020-02-11 23:34:39 发布

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探讨了一个关于整数的游戏，玩家选择一个大于1的正整数N，并通过不断随机选取该数的除数并相除的方式，直到N变为1。文章提供了计算达到目标所需的期望步数的算法。

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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
3
1
2
50
Sample Output
Case 1: 0
Case 2: 2.00
Case 3: 3.0333333333

就是正常的记忆化dfs…
无亮点

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
double dp[100001];
int bj[100001];
double dfs(int z)
{
    if (bj[z])return dp[z];
    bj[z] = 1;
    int tt = 2;
    double sd = 0;
    for (int a = 2; a*a <= z; a++)
    {
        if (z%a)continue;
        if (a*a == z)
        {
            tt++;
            sd += dfs(a);
            continue;
        }
        sd += dfs(a);
        sd += dfs(z / a);
        tt += 2;
    }
    sd += tt;
    return dp[z] = sd / (tt - 1);
}
int main()
{
    int T;
    cin >> T;
    int n,u=0;
    bj[1] = 1;
    while (T--)
    {
        cin >> n;
        double jg = dfs(n);
        printf("Case %d: %.7f\n", ++u, jg);
    }
}