题目链接：https://ac.nowcoder.com/acm/contest/993/B/
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

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 (1 <= N <= 20,000) has some height of Hi (1 <= Hi <= 10,000) and a total height summed across all N cows of S. The bookshelf has a height of B (1 <= B <= S < 2,000,000,007).
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.

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

解题思路

题意:找到一堆奶牛使得它们的身高和大于等于书架的高度，求奶牛的最小数目。
思路:因为只要超过就行了，所以就一直找最大的就行了，直到大于等于书架的高度。

Accepted Code:

#include <bits/stdc++.h>
using namespace std;
int spt[20005];
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++)
        scanf("%d", spt + i);
    sort(spt, spt + n);
    int r = n - 1, cnt = 0;
    while (cnt < m && ~r) {
        cnt += spt[r--];
    }
    printf("%d\n", n - r - 1);
    return 0;
}