Rating - HDU 4870 期望dp

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Rating

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 403 Accepted Submission(s): 254
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

1.0000000.814700

Sample Output

39.00000082.181160

题意：一个人有两个账号，每次用积分少的号比赛，有p的概率加50分，还有1-p的概率减100分，最低减到0，问有一个号到1000分的期望是多少。

思路：dp[i]表示从i分增加到i+1分的期望。dp[i]=p+(1-p)*(dp[i-2]+dp[i-1]+dp[i]+1)。整理后dp[i]=1/p+(1-p)/p*(dp[i-2]+dp[i-1]); dp[0]=1/p,dp[1]=1/p/p。然后就可以推出结果了。

AC代码如下：

#include<cstdio>#include<cstring>using namespace std;double dp[30],ans[30][30],p;int main(){ int i,j,k;  while(~scanf("%lf",&p))  { dp[0]=1/p;    dp[1]=1/p/p;    for(i=2;i<=19;i++)     dp[i]=1/p+(1-p)/p*(dp[i-2]+dp[i-1]);    for(i=0;i<=20;i++)    { ans[i+1][i]=ans[i][i]+dp[i];      ans[i+1][i+1]=ans[i+1][i]+dp[i];    }    printf("%.6f\n",ans[20][19]);  }}

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