LightOJ - 1038 Race to 1 Again —— 期望

题目链接：https://vjudge.net/problem/LightOJ-1038

 

1038 - Race to 1 Again

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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

 

 

题意：

给出一个数n，每次随机变为它的某一个因子(1~n)，直到最后变为1。问n平均多少步操作到达1？

 

题解：

1.假设ci为n的因子，那么：dp[n] = (dp[1]+1 + dp[c2]+1 + dp[c3]+1 + ……  + dp[n]+1)/k，k为因子个数。

2.由于左右两边都有dp[n]，那么移项得：dp[n] = (dp[1] + dp[c2] + dp[c3] + ……  + k)/(k-1) 。

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 1e5+100;

double dp[MAXN];
void init()
{
    memset(dp, 0, sizeof(dp));
    for(int i = 2; i<MAXN; i++)
    {
        int cnt = 0; double sum = 0;
        for(int j = 1; j<=sqrt(i); j++) if(i%j==0) {
            sum += dp[j];
            cnt++;
            if(j!=i/j) sum += dp[i/j], cnt++;
        }
        dp[i] = 1.0*(sum+cnt)/(cnt-1);
    }
}


int main()
{
    init();
    int T, n, kase = 0;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d",&n);
        printf("Case %d: %.10lf\n", ++kase, dp[n]);
    }
}