hdu 1145 So you want to be a 2n-aire?

题目地址：

http://acm.hdu.edu.cn/showproblem.php?pid=1145

题目描述：

So you want to be a 2n-aire?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 250    Accepted Submission(s): 188

Problem Description

The player starts with a prize of $1, and is asked a sequence of n questions. For each question, he may 
quit and keep his prize. 
answer the question. If wrong, he quits with nothing. If correct, the prize is doubled, and he continues with the next question. 
After the last question, he quits with his prize. The player wants to maximize his expected prize. 
Once each question is asked, the player is able to assess the probability p that he will be able to answer it. For each question, we assume that p is a random variable uniformly distributed over the range t .. 1. 

 

Input

Input is a number of lines, each with two numbers: an integer 1 ≤ n ≤ 30, and a real 0 ≤ t ≤ 1. Input is terminated by a line containing 0 0. This line should not be processed.

 

Output

For each input n and t, print the player's expected prize, if he plays the best strategy. Output should be rounded to three fractional digits.

 

Sample Input

   
   
   
   

    
    
    
    1 0.5
1 0.3
2 0.6
24 0.25
0 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1.500
1.357
2.560
230.138
   
   
   
   
   
   
   
   


   
   
   
   

题意：

一个人有1$钱，然后答n道题，可以不答带原有奖金离开，或者答题获得双倍奖金然后继续答题。问最佳收益是多少。

题解：

此题主要是有这样的决策，面对第i+1道题时到底是答还是不答，已经答完i道题的情况下。
明显答还是不答要看i道题的后一题i+1道题的期望收益，如果总体期望收益大于i道题的，我干吗不答呢？？
所以就有假设:
p*ex[i+1] > 2^i, 答
     左侧p表示i+1题答对的概率，ex[i+1]表示i+1题以及考虑后面题的最佳期望收益奖金（最佳是表示不仅考虑了i+1,还考虑了后面的i+2,i+3......,n的答题情况的综合期望，代码倒推循环的原因也在于此,所以ex[0]是综合了第1道题 到 第n道题的综合最佳收益奖金，所以我们最终输出ex[0]）
p*ex[i+1] < 2^i,不答  （我答了题奖金反而减少）

由此我们可以得出一个临界概率，bp=2^i/ex[i+1]

然后注意t值不是我们答题的概率，t只是限定我们答题概率的变化边界（一开始读题就误导t的含义了），我们答题的概率是[t,1)

边界概率bp < t: 本题的最佳收益是 ex[i] = (1+t)/2 * ex[i+1]
     说明随便答，p随意取[t,1)的值都是 p*ex[i+1] > 2^i 所以直接用平均概率去答题就ok了，注意这个平均概率这个平均是（起点 + 终点）/2 的意思，直接画个一维数轴就行了。也就是说答i题的收益是 p*ex[i+1] 而p我们取的是平均的概率即p=(1+t)/2

边界概率bp > t: 本题的最佳收益是 ex[i] = (bp-t)/(1-t) * 2^i + (1-bp)/(1-t) * (1+bp)/2 * ex[i+1]
     这里最好画个数轴要清楚点，当bp > t时 ，在数轴上 bp 肯定在[t,1)之间，那么[t,bp)表明不答i+1题，[bp,1)表明答i+1题,注意大分母是1-t,不是1，看下概率密度的概念，画个数轴就知道了。

代码：

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
using namespace std;
double ex[30+5]={0.0};//best expect
double mon[30+5]={0.0};//best money
double t=0.0;//probability p is t < p < 1
int n=0;//number of question
double bp=0.0;//boundary probability, if p > bp then get in the question  otherwise get out the question
int main()
{
    mon[0]=1;//after answer 0 question,the index begin with 1
    for(int i=1;i<=30;i++)
    {
        mon[i]=mon[i-1]*2;
    }
    while(scanf("%d%lf",&n,&t)!=EOF&&(n>0))
    {
        ex[n]=mon[n];
        //from ex[i+1] get the ex[i],ex[0] is our answers
        for(int i=n-1;i>=0;i--)
        {
            bp=mon[i]/ex[i+1];// 2^i/ex[i+1] is the boundary probability ==>  bp * ex[i+1] > 2^i
            if(bp<=t)//bp is not in [t,1) range, all p in [t,1) will let the p * ex[i+1] >2^i
            {
                ex[i]=(1+t)/2 * ex[i+1];
            }
            else// E = p*x + (1-p)*y
            {
                ex[i]=(bp-t)/(1-t) * mon[i] + (1-bp)/(1-t) * (1+bp)/2 * ex[i+1];
            }
        }
        printf("%.3lf\n",ex[0]);
    }
    return(0);
}