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+;
 const int MAXN = 1e5+;
 
 double dp[MAXN];
 void init()
 {
     memset(dp, , sizeof(dp));
     for(int i = ; i<MAXN; i++)
     {
         int cnt = ; double sum = ;
         for(int j = ; j<=sqrt(i); j++) if(i%j==) {
             sum += dp[j];
             cnt++;
             if(j!=i/j) sum += dp[i/j], cnt++;
         }
         dp[i] = 1.0*(sum+cnt)/(cnt-);
     }
 }
 
 int main()
 {
     init();
     int T, n, kase = ;
     scanf("%d", &T);
     while(T--)
     {
         scanf("%d",&n);
         printf("Case %d: %.10lf\n", ++kase, dp[n]);
     }
 }