HDU 4870 rating

Rating Time Limit: 10000/5000 MS (Java/others) memory limit: 32768/32768 K (Java/Others)
Total submission (s): 414 accepted submission (s): 261
Special Judge

Problem descriptiona little girl loves Programming Competition very much. recently, she has found a new kind of Programming Competition named "toptoptopcoder ". every user who has registered in "toptoptopcoder" system will have a rating, and the initial value of rating equals to zero. after the user participates in the contest held by "toptoptopcoder", her/his rating will be updated depending on her /His rank. supposing that her/his current rating is X, if her/his rank is between on 1-200 after contest, her/his rating will be min (x + 50 ). her/his rating will be max (X-100, 0) otherwise. to reach 1000 points as soon as possible, this little girl registered two accounts. she uses the account with less rating in each contest. the possibility of her rank between on 1-200 is P for every contest . Can you tell her how her account ratings reach 1000 points?
Inputthere are several test cases. Each test case is a single line containing a float number P (0.3 <= P <= 1.0). The meaning of P is described above.
Outputyou shoshould output a float number for each test case, indicating the expected Count of contest she needs to participate in. This problem is special judged. The relative error less than 1e-5 will be accepted.
Sample Input

1.0000000.814700

 
Sample output

39.00000082.181160

 
Authorfzu
Source2014 multi-university training contest 1
Question: When a girl plays a game, if the first 200 match results, she will be given a score of 50 for her rating. Otherwise, she will score 100 (the minimum rating value is 0, the maximum value is 1000 ---- the probability of being able to enter the first 200 is P ). In order to reach 1000 points, the girl used two accounts for the competition, each time using the account with low rating, until one account rating reached 1000. Given a P, the question is the expected value of the number of matches.

Question: The first thing we think of is the push formula, which uses DP [I] to represent the expected value from I * 50-(I + 1) * 50. DP [0] and DP [1] must be processed separately.

DP [0] indicates the number of fields we need to perform from 0 to 50, which is divided into two situations: 1. Success, probability is P, expected to be 1 * P

2. Failure, probability 1-P, expected to be (1-p) * (1 + dp [0])

----- Therefore, DP [0] = 1 * P + (1-p) * (1 + dp [0]). After simplification, DP [0] = 1/P;

DP [1] represents our expectation from the number of 50-fields, divided into two situations: 1. Success, probability is P, expectation is 1 * P

2. Failure, probability 1-P, expected to be (1-p) * (1 + dp [0] + dp [1])

----- So DP [1] = 1 * P + (1-p) * (1 + dp [0] + dp [1]), after simplification, DP [1] = 1 + (1-p)/P * (1 + dp [0]);

I> 2, the DP [I] method is divided into two situations: 1. Success, probability is P, expectation is 1 * P

2. Failure, probability 1-p, expected to be (1-p) * (1 + dp [I-2] + dp [I-1] + dp [I])

----- So DP [1] = 1 * P + (1-p) * (1 + dp [0] + dp [1]), after simplification DP [1] = 1 + (1-p)/P * (1 + dp [I-2] + dp [I-1]);

In this way, because we want to use two accounts for the competition, the final arrival status is one account rating = 1000, and the other is = 950. We only need to sum the DP.

* In addition, this question can also be done with Gaussian elimination. Currently, no, I am learning. I am waiting to add it.

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int main(){    double p,q;    double dp[20];    while(cin>>p)    {        memset(dp,0,sizeof(dp));        q=1.0-p;        double sum=0;        dp[0]=1.0/p;        dp[1]=1.0+q/p*(1+dp[0]);        sum+=(dp[0]+dp[1])*2;        for(int i=2;i<=19;i++)        {            dp[i]=1.0+q/p*(1+dp[i-1]+dp[i-2]);            sum+=dp[i]*2;        }        printf("%.6lf\n",sum-dp[19]);    }    return 0;}