HDOJ 4870 Rating

概率DP.....

Rating

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 229    Accepted Submission(s): 140
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
   
   
   
   

 

Author

FZU

 

Source

2014 Multi-University Training Contest 1

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

double p,dp[100],ans[100][100];

int main()
{
while(scanf("%lf",&p)!=EOF)
{
    memset(dp,0,sizeof(dp));
    memset(ans,0,sizeof(ans));
    dp[0]=1./p;
    dp[1]=1+(1-p)/p*(1+dp[0]);
    for(int i=2;i<=22;i++)
        dp[i]=1.+((1.-p)/p)*(1+dp[i-1]+dp[i-2]);
    ans[0][0]=0;
    ans[1][0]=dp[0]; ans[1][1]=ans[1][0]+dp[0];
    for(int i=2;i<=20;i++)
    {
        ans[i][i-1]=ans[i-1][i-1]+dp[i-1];
        ans[i][i]=ans[i][i-1]+dp[i-1];
    }
    printf("%.10lf\n",ans[20][19]);
}
    return 0;
}