题目描述

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 each with some height of - these are very tall cows). The bookshelf has a height of , 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.

输入描述:

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

输出描述:

* 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.

示例1

输入
5 16
3
1
3
5
6
输出
1
说明
Here we use cows 1, 3, 4, and 5, for a total height of 3 + 3 + 5 + 6 = 17.
It is not possible to obtain a total height of 16, so the answer is 1.

解答

1 确定可能的最大高度sum，就是所有的cow加起来的高度

2 根据动态规划法，求解1到最大高度sum之间的可能解

3 找到比B（书架高度）的最低高度，可能和B一致。 

#include <stdio.h>
#include <vector>
#include <limits.h>
#include <string.h>
#include <algorithm>
using namespace std;
 
const int MAX_N = 21, MAX_H = 1000000;
int cow[MAX_N];
bool height[MAX_N*MAX_H];
 
int getMinHeight(int N, int B, int sum)//B < sum
{
	fill(height, height+sum+1, false);
	height[0] = true;
	for (int i = 0; i < N; i++)
	{
		for (int j = sum; j >= cow[i]; j--)
		{
			if (height[j-cow[i]]) height[j] = true;
		}
	}
	int ans = B;
	for (; ans <= sum && !height[ans]; ans++) {}
 
	return ans;
}
 
int main()
{
	int N, B, sum;
	while (~scanf("%d %d", &N, &B))
	{
		sum = 0;
		for (int i = 0; i < N; i++)
		{
			scanf("%d", cow+i);
			sum += cow[i];
		}
		printf("%d\n", getMinHeight(N, B, sum)-B);
	}
	return 0;
}

来源：靖心