hdu 5781 ATM Mechine(2016 Multi-University Training Contest 5——DP)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5781

ATM Mechine

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 713    Accepted Submission(s): 332

Problem Description

Alice is going to take all her savings out of the ATM(Automatic Teller Machine). Alice forget how many deposit she has, and this strange ATM doesn't support query deposit. The only information Alice knows about her deposit is the upper bound is K RMB(that means Alice's deposit x is a random integer between 0 and K (inclusively)). 
Every time Alice can try to take some money y out of the ATM. if her deposit is not small than y, ATM will give Alice y RMB immediately. But if her deposit is small than y, Alice will receive a warning from the ATM. 
If Alice has been warning more then W times, she will be taken away by the police as a thief.
Alice hopes to operate as few times as possible.
As Alice is clever enough, she always take the best strategy. 
Please calculate the expectation times that Alice takes all her savings out of the ATM and goes home, and not be taken away by the police.

 

Input

The input contains multiple test cases.
Each test case contains two numbers K and W.
1≤K,W≤2000

 

Output

For each test case output the answer, rounded to 6 decimal places.

 

Sample Input

  

   1 1
4 2
20 3
  

 

Sample Output

  

   1.000000
2.400000
4.523810
  

 

Author

ZSTU

 

Source

2016 Multi-University Training Contest 5

 

题目大意：

你只知道在银行中0-k之间这么多的存款，然后你要把银行里的钱全部都取出来。

但是在取钱过程中，如果你取的钱超过了你在银行存的钱，那么你会被警告。

警告次数不可以超过w次，超过w就会被警察抓走。

问采用最优策略之后，取完所有钱的期望是多少？

解题思路：

dp[i][j]表示：存款数为[0,i]，还可以被警告j次。dp[i][j]表示最少要取多少次。

推导动态转移方程：

1、要是我这次取了钱k，但是没警告的话，就为dp[i-k][j]。

2、取钱警告了，方程为dp[k-1][j-1]。

这样方程为：dp[i][j]=min(dp[i][j],dp[i-k][j]+dp[k-1][j-1]+i+1);

详见代码。

#include <iostream>
#include <cstdio>

using namespace std;

#define INF 210000000
#define ll long long
ll dp[2010][21];

int main()
{
    for (int i=0;i<=2000;i++)
    {
        for (int j=0;j<=15;j++)
        {
            dp[i][j]=INF;
        }
    }
    for (int j=0;j<=15;j++)
    {
        dp[0][j]=0;
    }
    for (int i=1;i<=2000;i++)
    {
        for (int j=1;j<=15;j++)
        {
            for (int k=1;k<=i;k++)
            {
                dp[i][j]=min(dp[i][j],dp[i-k][j]+dp[k-1][j-1]+i+1);
            }
        }
    }
    int k,w;
    while (~scanf("%d%d",&k,&w))
    {
        if (w>15)
            w=15;
        ll a=dp[k][w];
        double ans=a*1.0/(k+1);
        printf ("%.6lf\n",ans);
    }
    return 0;
}