HDU 5781 ATM Mechine

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 14 220 3

 

Sample Output

1.0000002.4000004.523810
求期望，直接dp把答案都跑出来就好了，瞎猜了一波单调，貌似对了？
#include<set>#include<map>#include<cmath>#include<stack>#include<queue>#include<bitset>#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<functional>#define rep(i,j,k) for (int i = j; i <= k; i++)#define per(i,j,k) for (int i = j; i >= k; i--)using namespace std;typedef __int64 LL;const int low(int x) { return x&-x; }const double eps = 1e-8;const int INF = 0x7FFFFFFF;const int mod = 1e9 + 7;const int N = 2e3 + 10;int T, n, m;double dp[N + 1][N + 1];int main(){    rep(i, 0, N) dp[0][i] = 0;    rep(i, 1, N) dp[i][1] = dp[i - 1][1] * i / (i + 1) + 1;    rep(i, 1, N)    {        int k = 1;        rep(j, 2, N)        {            dp[i][j] = INF;            int kk = 1;            for (k; k <= i; k++)            {                if (dp[i][j] > (dp[i - k][j] * (i - k + 1) + dp[k - 1][j - 1] * k) / (i + 1))                {                    dp[i][j] = (dp[i - k][j] * (i - k + 1) + dp[k - 1][j - 1] * k) / (i + 1);                    kk = k;                }                if (kk < k) break;            }            k = kk;            dp[i][j] += 1;            //printf("%d %d\n", i, kk);            //int k = (i + 1) / 2;            //dp[i][j] = (dp[i - k][j] * (i - k + 1) + dp[k - 1][j - 1] * k) / (i + 1) + 1;        }    }    while (scanf("%d%d", &n, &m) != EOF)    {        printf("%.6lf\n", dp[n][m]);    }    return 0;}