题目描述

Farmer John recently bought a bookshelf for cow library, but the shelf is getting filled up quite quickly, and now the only available space is at the top.
Each of the N cows () has some height of and a total height summed across all N cows of S. The bookshelf has a height of .
To reach the top of the bookshelf taller than the tallest cow, 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. Since more cows than necessary in the stack can be dangerous, your job is to find the set of cows that produces a stack of the smallest number of cows possible such that the stack can

输入描述:

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

输出描述:

* Line 1: A single integer representing the size of the smallest set of cows that can reach the bookshelf.

示例1

输入
6 40
6
18
11
13
19
11
输出
3
说明
Six cows; bookshelf height 40. Cow heights fall in the range 6..19.
One way to reach 40 with 3 cows is 18+11+13; many others exist

解答

题意分析：

每个牛都有一个高度，一头牛可以站在其他牛上，这样它们的高度就是高度之和

应该是一群牛要上顶层，要把一部分牛堆起来，要能够上顶层，累加起来的高度必须大于给出的书架？高度，求最少的牛数目。

解题思路：

把牛从高到底排序，累加，大于书架高度时退出。 
#include <stdio.h>
#include <algorithm>
using namespace  std;
#define N 20020
int a[N];
int cmp(int a, int b) {
	return a > b ;
}
int main()
{
	int n, b, i, sum;
	scanf("%d%d", &n, &b);
	for (i = 0; i < n; i++)
		scanf("%d", &a[i]);
	sort(a, a + n, cmp);
	sum = 0;
	for (i = 0; i < n; i++)
	{
		sum += a[i];
		if (sum >= b)
			break;
	}
	printf("%d\n", i + 1);
	return 0;
}

来源：缘定三生石