LightOJ 1038 Race to 1 Again（概率DP）

最新推荐文章于 2021-06-22 21:54:43 发布

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最新推荐文章于 2021-06-22 21:54:43 发布

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本文链接：https://blog.youkuaiyun.com/qq_41746268/article/details/90244254

探讨了如何使用概率DP解决一个数学问题，即给定一个正整数N，求将其通过连续除以其因子的方式变回1所需的平均次数。通过建立状态转移方程并运用动态规划，最终求得N变回1的期望次数。

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Race to 1 Again

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

题意
对于一个数 $n$ ，求把它变回1，所需要的除的次数的期望

题解
概率DP。
令 $d p [i]$ 表示需要次数为 $i$ 的期望
状态转移方程：
$\begin{aligned} dp[i]&=\cfrac{\sum_{d|i}{dp[d]}}{num[i]}+1\\ dp[i]num[i]&=\sum_{d|i}dp[d]+num[i]\\ dp[i](num[i]-1)&=\sum_{d|i,d=\not i}dp[d]+num[i]\\ dp[i]&=\cfrac{\sum_{d|i,d=\not i}dp[d]+num[i]}{num[i]-1} \end{aligned}$
其中 $d ∣ i$ 表示 $d$ 是 $i$ 的因子， $n u m [i]$ 表示 $i$ 的因子个数

求解时，对每个 $i (i > 1)$ 都通过公式求出其 $d p [i]$ ,并对 $n u m [i] + 1$ （ $i$ 本身也是因子），然后对 $i$ 的倍数 $j$ 都进行 $d p [j] + = d p [i]$ 并且 $n u m [j] + 1$ ,
代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <queue>
#include <cmath>
#include <string>
#include <cstring>
#include <map>
#include <set>
#include <cmath>

using namespace std;
#define me(x,y) memset(x,y,sizeof x)
#define MIN(x,y) x < y ? x : y
#define MAX(x,y) x > y ? x : y
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 1e5;
const ll INF = 0x3f3f3f3f;
const int MOD = 998244353;
const double eps = 1e-06;

double dp[maxn+10];
int num[maxn+10];
void get_dp(){
    me(dp,0);
    me(num,0);
    dp[1] = 0;
    for(int i = 1; i <= maxn; ++i){
        num[i]++;
        if(i>1)dp[i] = (dp[i]+num[i])/(num[i]-1);
        for(int j = i*2; j <= maxn; j += i){
            dp[j] += dp[i];
            num[j]++;
        }
    }
}

int main(){
    int t;
    get_dp();
    cin>>t;
    for(int ca = 1; ca <= t; ca++){
        int n;
        cin>>n;
        printf("Case %d: %lf\n",ca,dp[n]);
    }
    return 0;
}


/*


*/