Rating

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 786    Accepted Submission(s): 496
Special Judge 

Problem Description

A 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+50,1000). 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 many contests she needs to participate in to make one of her account ratings reach 1000 points?

 

Input

There 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.

 

Output

You should 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

1.000000

0.814700

 

Sample Output

39.000000

82.181160

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题意 ：小女孩注册了两个比赛的帐号，初始分值都为0，每做一次比赛如果排名在前两百名，rating涨50，否则降100，告诉你她每次比赛在前两百名的概率p，如果她每次做题都用两个账号中分数低的那个去做，问她最终有一个账号达到1000分需要做的比赛的次数的期望值。

思路 ：可以直接用公式推出来用DP做，也可以列出210个方程组用高斯消元去做。

（1）DP1：离散化。因为50，100，1000都是50的倍数，所以就看作1，2，20。这样做起来比较方便。

定义dp[i]为从 i 分数到达i+1分的期望，状态转移方程：

 dp[i] = p+(1-p)*(1+dp[i-2]+dp[i-1]+dp[i]); 在前两百名里增加一分，当不在前两百名里的时候，扣两分，要回到 i+1 分就是1+dp[i-2]+dp[i-1]+dp[i].

mp[i][i]表示两个账号都从0分涨到 i 分的期望，所以mp[i+1][i] = mp[i][i]+dp[i], mp[i+1][i+1] = mp[i+1][i]+dp[i];

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>

using namespace std ;

double dp[21],mp[21][21] ;

int main()
{
    double p ;
    while(scanf("%lf",&p) != EOF)
    {
        dp[0] = 1 / p ;
        dp[1] = 1 / p / p ;
        for(int i = 2 ; i < 20 ; i++)
            dp[i] = 1 + (1-p)*(dp[i-2]+dp[i-1]+1)/p ;
        for(int i = 0 ; i < 20 ; i++)
        {
            mp[i+1][i] = mp[i][i]+dp[i] ;
            mp[i+1][i+1] = mp[i+1][i] + dp[i] ;
        }
        printf("%.6lf
",mp[20][19]) ;
    }
    return 0 ;
}