首页 > 代码库 > HDOJ 4870 Rating
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; }
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。