poj 3268 Bookshelf 2

/*Bookshelf 2
Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9314		Accepted: 4215
Description

Farmer John recently bought another bookshelf for the cow library, but the shelf is getting 
filled up quite quickly, and now the only available space is at the top.

FJ has N cows (1 ≤ N ≤ 20) each with some height of Hi (1 ≤ Hi ≤ 1,000,000 - these are 
very tall cows). The bookshelf has a height of B (1 ≤ B ≤ S, where S is the sum of the heights 
of all cows).

To reach the top of the bookshelf, one or more of the cows can stand on top of each other in a
 stack, so that their total height is the sum of each of their individual heights. This total
  height must be no less than the height of the bookshelf in order for the cows to reach the top.

Since a taller stack of cows than necessary can be dangerous, your job is to find the set of 
cows that produces a stack of the smallest height possible such that the stack can reach the 
bookshelf. Your program should print the minimal 'excess' height between the optimal stack of
 cows and the bookshelf.

Input

* Line 1: Two space-separated integers: N and B
* Lines 2..N+1: Line i+1 contains a single integer: Hi

Output

* Line 1: A single integer representing the (non-negative) difference between the total 
height of the optimal set of cows and the height of the shelf.

Sample Input

5 16
3
1
3
5
6
Sample Output

1
就是给出n和B，然后给出n个数，用这n个数中的某些，求出一个和，这个和是>=B的最小值，
输出最小值与B的差。*/
#include 
#include
#define max(a,b) a>b?a:b
#define min(a,b)a=h[i];j--)
			{
					dp[j]=max(dp[j],dp[j-h[i]]+h[i]);
					if(dp[j]>=B&&dp[j]