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
题意：有N 头牛，书架高度M， 列举每头牛的高度， 求牛叠加起来的超过书架的最小高度。
题目思路：求出奶牛叠加能达到的所有高度，并用dp[]保存，最后进行遍历，找出与h差最小的dp[]即所求答案。
代码：
#include<stdio.h>
#include<string.h>
#define max(a, b)(a > b ? a : b)
#define N 1000010
int main(void)
{
int dp[N];
int i, j, n, m;
int w[N];
while(scanf("%d%d", &n, &m) != EOF)
{
int sum = 0;//所有牛加起来的总高度。
memset(dp, 0, sizeof(dp));
for(i = 1; i <= n; i++)
{
scanf("%d", &w[i]);
sum += w[i];
}
int ans = 0;
for(i = 1; i <= n; i++)
{
for(j = sum; j >= w[i]; j--)//j要倒推才能保证在推dp[j]时， max里dp[j]和dp[j-w[i]]保存的是状态dp[i-1][j] 和dp[i-1][j-w[i]]的值。
{
dp[j] = max(dp[j], dp[j-w[i]] + w[i]);
}
}
for(i = 1; i <= sum; i++)//遍历dp数组找到超过书架的最小高度直接保存跳出循环。
{
if(dp[i] >= m)
{
ans = dp[i];
break;
}
}
printf("%d ", ans - m);
}
return 0;
}