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 H_i (1 ≤ H_i ≤ 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.

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

* 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.

5 16
3
1
3
5
6

1
简单的 0，1背包问题
01背包（ZeroOnePack）: 有N件物品和一个容量为V的背包。（每种物品均只有一件）第i件物品的费用是c[i]，价值是w[i]。求解将哪些物品装入背包可使价值总和最大。

01背包（ZeroOnePack）: 有N件物品和一个容量为V的背包。（每种物品均只有一件）第i件物品的费用是c[i]，价值是w[i]。求解将哪些物品装入背包可使价值总和最大。

这是最基础的背包问题，特点是：每种物品仅有一件，可以选择放或不放。

用子问题定义状态：即f[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。则其状态转移方程便是：

f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}

把这个过程理解下：在前i件物品放进容量v的背包时，

它有两种情况：

第一种是第i件不放进去，这时所得价值为:f[i-1][v]

第二种是第i件放进去，这时所得价值为：f[i-1][v-c[i]]+w[i]

（第二种是什么意思？就是如果第i件放进去，那么在容量v-c[i]里就要放进前i-1件物品）

最后比较第一种与第二种所得价值的大小，哪种相对大，f[i][v]的值就是哪种。

（这是基础，要理解！）

这里是用二位数组存储的，可以把空间优化，用一位数组存储。

用f[0..v]表示，f[v]表示把前i件物品放入容量为v的背包里得到的价值。把i从1~n(n件)循环后，最后f[v]表示所求最大值。

*这里f[v]就相当于二位数组的f[i][v]。那么，如何得到f[i-1][v]和f[i-1][v-c[i]]+w[i]？（重点！思考） 首先要知道，我们是通过i从1到n的循环来依次表示前i件物品存入的状态。即：for i=1..N 现在思考如何能在是f[v]表示当前状态是容量为v的背包所得价值，而又使f[v]和f[v-c[i]]+w[i]标签前一状态的价值？

逆序！

这就是关键！

分析上面的代码：当内循环是逆序时，就可以保证后一个f[v]和f[v-c[i]]+w[i]是前一状态的！ 此题的代码

#include <stdio.h>
#include <string.h>int dp[1000005],a[10005];int max(int a,int b)
{return a>b?a:b;
}int main()
{int n,m;while(~scanf("%d%d",&n,&m)){int i,j,sum = 0;memset(dp,0,sizeof(dp));memset(a,0,sizeof(a));for(i = 1; i<=n; i++){scanf("%d",&a[i]);sum+=a[i];}for(i = 1; i<=n; i++){for(j = sum; j>=a[i]; j--)dp[j] = max(dp[j],dp[j-a[i]]+a[i]);}for(i = 1; i<=sum; i++){if(dp[i]>=m){printf("%d\n",dp[i]-m);break;}}}return 0;
}