HDU 5781 ATM Mechine (期望DP)

题意：你有一些钱数量在 0 ~ n 之间，然后你要取钱，但是如果取的钱数超过你的钱数，会被警告，问警告不超过m次，把钱取走的期望是多少。

析：dp[i][j]  表示钱在 0 ~ i，然后最多 j 次警告，根据二分的思想，j 最大是11，然后每次取 k 元，分为两种情况，一种是钱数不小于k，那么就会被警告，就会转移到 dp[k-1][j-1]，如果线数不小于k，那么转移到 dp[i-k][j]。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2000 + 10;
const LL mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
  return r > 0 && r <= n && c > 0 && c <= m;
}

int dp[maxn][15];

void init(){
  ms(dp, INF);  ms(dp[0], 0);
  for(int i = 1; i < maxn; ++i)
    for(int j = 1; j < 12; ++j)
      for(int k = 1; k <= i; ++k)
        dp[i][j]= min(dp[i][j], dp[i-k][j] + dp[k-1][j-1] + 1 + i);
}

int main(){
  init();
  while(scanf("%d %d", &n, &m) == 2)
    printf("%.6f\n", dp[n][min(11, m)] / (n + 1.0));

  return 0;
}