hdu-5781 ATM Mechine(dp+概率期望)

題目鏈接：

ATM Mechine

Time Limit: 6000/3000 MS (Java/Others)???

?Memory Limit: 65536/65536 K (Java/Others)

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

題意：

問等概率為[0,k]中的一個,最多被警告w次,問取錢次數的最小期望是多少;

思路：

題解講的很清楚了,就是dp啦,二分可知警告次數不超過15次,每次選一個數取,能取出來和不能取出來兩種情況，加在一起就是題解的式子了:

dp[i][j]=min( dp[i-k][j]*(i-k+1)/(i+1)+dp[k-1][j-1]*k/(i+1)+1 )

邊界就是dp[0][i]=0;其他的要正無窮;

AC代碼：

/************************************************
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┆　　┃　 AC代馬 　　┣┓┆
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┆　　　┃┫┫　┃┫┫ ┆
┆　　　┗┻┛　┗┻┛ ┆      
************************************************ */ #include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>using namespace std;#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));typedef  long long LL;template<class T> void read(T&num) {char CH; bool F=false;for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {if(!p) { puts("0"); return; }while(p) stk[++ tp] = p%10, p/=10;while(tp) putchar(stk[tp--] + '0');putchar('\n');
}const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e6+10;
const int maxn=2e3+14;
const double eps=1e-8;double dp[maxn][17];inline void Init()
{For(i,1,maxn-5)For(j,0,16)dp[i][j]=inf;For(j,1,16)dp[0][j]=0;For(i,1,maxn-5){For(j,1,16){For(x,1,i){dp[i][j]=min(dp[i][j],dp[i-x][j]*(i-x+1)/(i+1)+dp[x-1][j-1]*x/(i+1)+1);}}}}
int main()
{      Init();int k,w;while(cin>>k>>w){printf("%.6lf\n",dp[k][min(w,15)]);}return 0;
}