CodeForces 388A Fox and Box Accumulation

本文介绍了一个有趣的算法问题:如何用最少的堆叠次数将不同承重能力的盒子堆叠起来。通过排序和贪心策略,文章提供了一种解决方法。

A. Fox and Box Accumulation
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most xi boxes on its top (we'll call xi the strength of the box). 

Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.

Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than xi boxes on the top of i-th box. What is the minimal number of piles she needs to construct?

Input

The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x1, x2, ..., xn (0 ≤ xi ≤ 100).

Output

Output a single integer — the minimal possible number of piles.

Sample test(s)
input
3
0 0 10
output
2
input
5
0 1 2 3 4
output
1
input
4
0 0 0 0
output
4
input
9
0 1 0 2 0 1 1 2 10
output
3
Note

In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.

In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).


先排序,再从上往下贪心,就酱╮(╯▽╰)╭

虽然给定引用中未直接提及“Kuroni and Simple Strings”题目的详细信息,但通常这类题目可能与字符串处理、括号匹配等相关。一般而言,题目可能会给出一个由括号组成的字符串,要求找出能移除的最大数量的不相交的合法括号对,并输出移除这些括号对后的相关信息。 ### 解法分析 #### 栈解法 栈解法是处理括号匹配问题的经典方法。通过遍历字符串,将左括号压入栈中,遇到右括号时,若栈顶为左括号,则将栈顶元素弹出,表示这是一对匹配的括号。 ```python s = input() stack = [] pairs = [] for i, char in enumerate(s): if char == '(': stack.append(i) else: if stack: left_index = stack.pop() pairs.append((left_index + 1, i + 1)) if not pairs: print(0) else: print(1) print(len(pairs) * 2) result = [] for l, r in pairs: result.extend([l, r]) result.sort() print(" ".join(map(str, result))) ``` #### 双指针解法 双指针解法从字符串的两端向中间遍历,分别使用两个指针 `left` 和 `right`。`left` 指针从左向右寻找 `(`,`right` 指针从右向左寻找 `)`,当找到一对匹配的括号时,将它们标记为已移除,继续寻找下一对匹配的括号,直到无法再找到匹配的括号为止。 ```python s = input() n = len(s) left = 0 right = n - 1 pairs = [] while left < right: while left < right and s[left] != '(': left += 1 while left < right and s[right] != ')': right -= 1 if left < right: pairs.append((left + 1, right + 1)) left += 1 right -= 1 if not pairs: print(0) else: print(1) print(len(pairs) * 2) result = [] for l, r in pairs: result.extend([l, r]) result.sort() print(" ".join(map(str, result))) ``` ### 复杂度分析 - **栈解法**:时间复杂度为 $O(n)$,其中 $n$ 是字符串的长度。空间复杂度为 $O(n)$,主要用于栈的空间开销。 - **双指针解法**:时间复杂度为 $O(n)$,空间复杂度为 $O(n)$,主要用于存储匹配的括号对。
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值