hdu4870 Rating---2014 Multi-University Training Contest 1

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4870

Rating

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 171    Accepted Submission(s): 91
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

 

Author

FZU

 

Source

2014 Multi-University Training Contest 1

 

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大神们都说这题，应该是高斯消元才对，可惜。。屌丝这方面掌握得实在是太渣了。。。还是没能理解。。。

最后还是看了神牛们神一样的DP思路，才终于理解。。。哎。。惭愧。。。

设dp[i]为从i到i+1所需要的步数则有

                         dp[i]=p*1+(1-p)*(1+dp[i-2]+dp[i-1]+dp[i])  化简一下得 dp[i]=1+(1-p)/p*(1+dp[i-2]+dp[i-1])

                         又由已知条件可得 dp[0]=p*1+(1-p)*(1+dp[0]) 即 dp[0]=1/p

                        dp[1]=p*1+(1-p)*(1+dp[0]]+dp[1]) 即 dp[1]=1/p+(1-p)/p*(1+dp[0])

设ans[i][j]表示从初状态到两个号的分数为i，j的期望。两个号的分数的变化总是[i,i]->[i+1,i]->[i+1,i+1]；每次只有一个号的分数在进行变动。

所以有 ans[i+1][i] = ans[i][i]+dp[i], ans[i+1][i+1] = ans[i+1][i]+dp[i];

总的dp方程就是  ： dp[i]=p*1+(1-p)*(1+dp[i-2]+dp[i-1]+dp[i])     ans[i+1][i] = ans[i][i]+dp[i], ans[i+1][i+1] = ans[i+1][i]+dp[i];

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