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最新推荐文章于 2021-10-15 21:35:41 发布

转载最新推荐文章于 2021-10-15 21:35:41 发布 · 54 阅读

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原文链接：http://www.cnblogs.com/yxwkf/p/4820863.html

本文详细介绍了如何通过优化参赛策略，利用两次尝试机会，快速达到编程竞赛中特定积分的目标值。通过数学建模和期望值计算，揭示了在给定成功率的情况下，实现目标积分所需的期望比赛场数。本文不仅提供了理论分析，还给出了易于理解的代码实现，旨在帮助参赛者提高效率，更快地提升自己的排名。

Rating

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 414 Accepted Submission(s): 261
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

题目：一个女孩打比赛，每次比赛结果若在前200名则能给她的rating加上50分，否则将会将去100分（rating最小为0，最大为1000----可以进入前200的概率为p）。为了可以达到1000分。这个女孩使用两个帐号进行比赛，每次使用rating低的那个帐号比赛，直到有一个帐号rating达到1000。给定一个p。问最后须要进行比赛场数的期望值。

题解：首先我们想到的是推公式，以dp[i]代表从i*50-（i+1）*50的期望值。dp[0]和dp[1]须要单独处理。

dp[0]代表我们从0-50须要进行的场数，分成两种情况：1.成功，概率为p，期望为1*p

2.失败。概率1-p。期望为(1-p)*(1+dp[0])

-----所以dp[0]=1*p+(1-p)*(1+dp[0]) ，化简后dp[0]=1/p;

dp[1]代表我们从50-100的场数期望，分成两种情况：1.成功，概率为p，期望为1*p

2.失败，概率1-p，期望为(1-p)*(1+dp[0]+dp[1])

-----所以dp[1]=1*p+(1-p)*(1+dp[0]+dp[1]) ，化简后dp[1]=1+(1-p)/p*(1+dp[0]);

i>2,dp[i]的求法，分成两种情况：1.成功，概率为p。期望为1*p

2.失败，概率1-p，期望为(1-p)*(1+dp[i-2]+dp[i-1]+dp[i])

-----所以dp[i]=1*p+(1-p)*(1+dp[i-2]+dp[i-1]+dp[i]) 。化简后dp[i]=1+(1-p)/p*(1+dp[i-2]+dp[i-1]);

这样，由于要使用两个帐号进行比赛，所以我们最后到达的状态就是一个帐号rating=1000，另外一个=950，仅仅须要进行dp求和即可了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;

int main()
{
    double p,q;
    double dp[20];
    while(cin>>p)
    {
        memset(dp,0,sizeof(dp));
        q=1.0-p;

        double sum=0;
        dp[0]=1.0/p;
        dp[1]=1.0+q/p*(1+dp[0]);
        sum+=(dp[0]+dp[1])*2;

        for(int i=2;i<=19;i++)
        {
            dp[i]=1.0+q/p*(1+dp[i-1]+dp[i-2]);
            sum+=dp[i]*2;
        }
        printf("%.6lf\n",sum-dp[19]);
    }
    return 0;
}

高斯消元的做法（公式同上）：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const double eps=1e-11;  //设置为1e-9时Wa了
double a[40][40],x[40];
int equ,var;

void Debug()
{
    for(int i=0; i<equ; i++)
    {
        for(int j=0; j<=var; j++)
            printf("%4.2lf ",a[i][j]);
        puts("");
    }
    puts("");
}

double Gauss()
{
    int k,col,max_r;

    for(k=0,col=0; k<equ&&col<var; k++,col++)
    {
        max_r=k;
        for(int i=k+1; i<equ; i++)
            if(fabs(a[i][col])>fabs(a[max_r][col])) max_r=i;

        if(max_r!=k)
            for(int i=0; i<=var; i++)
            {
                swap(a[max_r][i],a[k][i]);
            }

        if(fabs(a[k][col])<eps)
        {
            k--;
            continue;
        }

        for(int i=k+1; i<equ; i++)
        {
            if(fabs(a[i][col])>eps)
            {
                double t=a[i][col]/a[k][col];
                for(int j=col; j<=var; j++)
                {
                    a[i][j]=a[i][j]-a[k][j]*t;
                }
            }
        }
    }

    //Debug();

    double ans=0;
    for(int j=var-1; j>=0; j--)
    {
        double temp=a[j][var];
        for(int r=j+1; r<var; r++)
            temp=(temp-a[j][r]*x[r]);

        x[j]=temp/a[j][j];
    }
    for(int j=0; j<var; j++)
        ans+=2.0*x[j];
    return ans-x[19];
}

void init(double p)
{
    equ=var=20;
    memset(a,0,sizeof(a));
    memset(x,0,sizeof(x));
    for(int i=0; i<20; i++)
    {
        a[i][var]=1.0;
    }
    a[0][0]=p;
    a[1][0]=p-1;
    a[1][1]=p;
    for(int i=2; i<=19; i++)
    {
        a[i][i]=p;
        a[i][i-1]=p-1;
        a[i][i-2]=p-1;
    }
    //Debug();
}

int main()
{
    double p;
    while(scanf("%lf",&p)!=EOF)
    {
        init(p);
        printf("%.6lf\n",Gauss());
    }
    return 0;
}

转载于:https://www.cnblogs.com/yxwkf/p/4820863.html