分类： 赛事动态

Ana is playing the game "pick-up sticks."安娜正在玩“捡棍子”的游戏。

She drops some sticks on a table and then picks them all up following these rules:她把一些木棍放在桌子上，然后按照以下规则把它们全部捡起来：

- You may pick up one stick at a time;你可以一次捡一根棍子；

- You may pick up a stick only if it is on top: no other stick may be covering it.只有当某根木棍在最上面，即没有其他木棍压在它上面时，你才可以把它捡起来。

Example:例子:

3 sticks are dropped and look like this. Press play to see how Ana picks them up in order:三根棍子掉落下来，排列成这样。点击播放，看看安娜是如何依次捡起它们的：

 

 

Question:问题：

Ana dropped 6 sticks like this:安娜像这样扔下了 6 根木棍：

 

 

 

 

 

Explanation 解释

Monica and Michael brought necklaces back from their vacation.莫妮卡和迈克尔度假回来时带回了项链。

Monica's necklace: 莫妮卡的项链:

Michael's necklace: 迈克尔的项链:

They each took several pearls off of their own necklace to make a new necklace for their friend Anastasia. Now their necklaces look like this. 他们各自从自己的项链上取下几颗珍珠，为他们的朋友安娜斯塔西娅做了一条新项链。现在他们的项链是这个样子。

Question:问题：

What does Anastasia's necklace look like?安纳斯塔西娅的项链是什么样子的？

 

 

 

 

 

Explanation 解释

Answer: 答案:

Monica's necklace is missing three violet pearls and Michael's is missing three red pearls, and that is exactly what is in this necklace.

莫妮卡的项链少了三颗紫珍珠，迈克尔的项链少了三颗红珍珠，而这串项链上恰好就有这些。is not correct because it contains four violet pearls. Monika’s necklace started with five violet pearls and she now has two left, so the maximum number that can be on Anastasija’s necklace is three.

这是不正确的，因为这条项链上有四颗紫珍珠。莫妮卡的项链起初有五颗紫珍珠，现在还剩两颗，所以安纳斯塔西娅的项链上最多只能有三颗。

is not correct because it contains blue and green pearls. All the green and blue pearls remain on Monika's and Michael's necklaces, so Anastasija’s necklace cannot have any green or blue pearls.

这是不正确的，因为其中包含蓝色和绿色的珍珠。所有绿色和蓝色的珍珠都还在莫妮卡和迈克尔的项链上，所以安娜斯塔西娅的项链上不能有任何绿色或蓝色的珍珠。

For the same reason, answer  is not correct.

出于同样的原因，这个答案不正确。

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Brian’s favorite candies come in five flavors. Brian puts one of each flavor into a tube to take to school. During the day, Brian eats the candies in the order they come out of the top of the tube. Today he wants to eat them in the following order: , , , , .

布莱恩最喜欢的糖果有五种口味。布莱恩把每种口味的一颗糖果放进一个管子里带去学校。白天，布莱恩按照糖果从管子顶部出来的顺序吃糖果。今天他想按照以下顺序吃：   。

Question 问题

Pack Brian’s tube for him so that he gets the candies in the preferred order.把布赖恩的管子给他装好，这样他就能按喜欢的顺序吃到糖果了。

(Drag the candies into the tube and press 'Save' when complete.)（将糖果拖入管中，完成后按“保存”。）

 

 

 

 

 

Explanation 解释

For the candies to come out of the tube in the preferred order, it is important to understand that the first one in will be the last one out. This means that the tube of sweets needs to be filled like this:为了让糖果按照理想的顺序从管子里出来，重要的是要明白先进入管子的糖果会最后出来。这意味着装糖果的管子需要这样装填：

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（Bebras每日一练部分展示）

Buffy has a pile of six clean shirts on her bed. She wants to put them away in her four drawers.巴菲在床上放着一堆六件干净的衬衫。她想把它们收进她的四个抽屉里。

She takes one shirt at a time from the pile and puts it into a drawer. Buffy starts with the top drawer, then uses the second from the top, and works her way down. When she has put a shirt into the bottom drawer, she starts again from the top.她每次从那堆衬衫里拿一件放进抽屉里。布菲从最上面的抽屉开始，接着用从上往下数第二个抽屉，然后依次往下放。当她把一件衬衫放进最下面的抽屉后，又从最上面的抽屉开始放。

 

 

 

 

 

 

 

 

Three children often borrow books from the library.三个孩子经常从图书馆借书。

The librarian wants to know which book has been borrowed most often. 图书管理员想知道哪本书被借阅的次数最多。

She uses the shapes on each child's library card to make a list of who borrowed which book.她根据每个孩子借书证上的形状来列出谁借了哪本书。

Task: 任务:

Look at her list in the middle. Which book was borrowed most often?看看中间她的清单。哪本书借阅次数最多？

(Click on the correct book on the right and press 'Save.')（点击右侧正确的书本，然后按“保存”。）

 

 

 

 

 

Explanation 解释

Task: 任务:

Paint the picture below. 请画出下面这幅画。

Painting Rules:绘画规则:

1. No part of the picture may be left white.1. 画面中任何部分都不许留白。

2. Parts that are touching cannot be the same color. For example, the leaf and the background may not both be green.2. 相接触的部分不能是相同的颜色。例如，叶子和背景不能都是绿色的。

Painting instructions:绘画说明：

Click on a paint color to choose it.点击一种油漆颜色以选择它。

Click on a part of the picture (like the background) to paint it.点击图片的某一部分（比如背景）来给它上色。

There is more than one way to correctly color the picture.给这幅画上色的方法不止一种。

Press 'Save' when you are finished.完成后请按“保存”。

 

 

 

 

Explanation 解释

Answer: 答:There are many possible answers, here is one:答案有很多，这里给出一个：Explanation: 解释:

The main rule in this task is: No painted areas that are next to each other are allowed to have the same colour.此任务的主要规则是：相邻的涂色区域不允许使用相同的颜色。

Start by chosing one of the three paints and paint the centre of the flower.首先从三种颜料中选择一种，给花朵的中心上色。

Next choose a different colour and paint the corner petals with it. They do not touch each other so can be the same colour.接下来选择一种不同的颜色，用它来给角落的花瓣上色。这些花瓣彼此不接触，所以可以是同一种颜色。

Next choose the third paint and colour in the remaining four petals. These also do not touch each other.接下来选择第三种颜料，给剩下的四片花瓣上色。这些花瓣之间也不要相互接触。

The middle of the flowerhead touches all the petals so must be a differnet colour to all the petals. The same applies to the background so this must be the same colour as the centre of the flower.花心部分与所有花瓣相接，所以颜色必须与所有花瓣不同。背景也是如此，所以其颜色必须与花心相同。

Finally, colour the stem and leaf while still obeying the main task rule.最后，给茎和叶上色，同时仍要遵循主要任务规则。

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2015年Bebras挑战练习题

Senior (16-18岁) 集合

名称	类型
Beaver Tutorials（海狸教程）	A
You Won't Find It（你不会找到的）	A
Drawing Stars（画星星）	A
Stack Computer（堆栈的电脑）	A
E-mail Scam（电子邮件诈骗）	A
Popularity（流行）	B
Beaver the Alchemist（炼金术士海狸）	B
Fireworks（烟花）	B
Bowl Factory（碗的工厂）	B
Word Chains（文字链）	B
Kangaroo（袋鼠）	C
Spies（间谍）	C
Mobiles（手机）	C
Pirate Hunters（海盗猎人）	C
Building a chip（构建芯片）	C

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（Bebras每日一练部分展示）

A small chip is composed of a grid of contacts (marked as dots). Some are already connected (marked as line segments). Connectors are always only between adjacent contacts, horizontally or vertically. We want to connect S and R with a continuous sequence of connectors, which do not touch any already connected contacts.

一个小芯片由一个触点网格（标记为点）组成。其中一些触点已经连接（标记为线段）。连接器总是仅在相邻触点之间水平或垂直连接。我们希望用一连串的连接器将 S 和 R 连接起来，且这些连接器不能接触任何已连接的触点。

Question: 问题:

How many different ways are there to connect S and R with the least possible number of connectors?

用最少的连接器将 S 和 R 连接起来，有多少种不同的连接方式？

A. 5

B. 13 

C. 15

D. 16

 

 

 

 

 

 

Explanation 解释

The correct answer is: 15

正确答案是：15

It is not difficult to find some shortest path (sequence of connectors). You can imagine a wave spreading over the board, one contact at a time, starting at S and moving toward R. When the wave reaches a contact, it certainly took the least possible number of connectors (“steps” of the wave).

要找到一些最短路径（连接器序列）并不难。你可以想象一个波浪在电路板上逐个接触点地扩散开来，从 S 点开始向 R 点移动。当波浪到达某个接触点时，它肯定已经经过了最少数量的连接器（波浪的“步数”）。

The table shows the wave midway through filling in. The numbers show the order in which the wave reached them, and the blacked out cells indicate the connectors we cannot connect to. They are also the length of the shortest path to the respective contact. The highest numbers are the current edge of the wave.

该表展示了波形在填充过程中的中间状态。数字表示波形到达各单元格的顺序，黑色单元格表示无法连接的连接器，它们也是到达相应触点的最短路径长度。最大的数字表示当前波形的前沿。

But if you try to find all the shortest paths, you can easily get lost. Luckily, you do not really need to, you are just interested in the number of paths. So how to break this task down to some smaller pieces?

但如果你试图找出所有最短路径，很容易迷失方向。幸运的是，你其实并不需要这么做，你只是对路径的数量感兴趣。那么如何将这个任务分解成一些更小的部分呢？

Realize this: A shortest path to any given contact must go through one of the adjacent contacts, which is exactly one connector closer to the start. If there are more such contacts, any of them can be used. So if you want to know the number of possibilities, you need to sum the possibilities to get to these adjacent contacts. The number of shortest paths to a certain contact is the sum of the numbers of the shortest paths to the adjacent contacts which are one connector closer to the start.

要明白这一点：到达任何给定接触点的最短路径必须经过其中一个相邻接触点，而该相邻接触点恰好比起始点更近一个连接器。如果有多个这样的接触点，那么其中任何一个都可以使用。所以，如果您想知道可能性的数量，就需要将到达这些相邻接触点的可能性相加。到达某个接触点的最短路径数量，就是到达比起始点更近一个连接器的相邻接触点的最短路径数量之和。

This allows you to follow a procedure, which will reliably give you the final number of shortest paths. You can proceed similarly as with finding the shortest path, filling in the table like a wave. Start at S. The contacts next to it can be reached in only one way. Then add the next contacts along the wave edge and always sum up the neighbour numbers from the previous step.

这使您能够遵循一个程序，该程序能可靠地得出最短路径的最终数量。您可以像寻找最短路径那样进行操作，像波浪一样逐步填满表格。从 S 开始。与它相邻的节点只能通过一种方式到达。然后沿着波浪边缘添加下一个节点，并始终将前一步中相邻节点的数量相加。

The resulting table looks like this, where we have also shown the order in which the cells are filled in by way of the table on the right, which is the completely filled in “shortest path” table discussed above:

最终得到的表格如下所示，我们还在右侧的表格中展示了单元格的填充顺序，该表格即为上文所讨论的已完全填好的“最短路径”表：

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（Bebras每日一练部分展示）

A mobile is a piece of art that hangs from a ceiling. You may remember one hanging from the ceiling in your bedroom.

风铃是一种悬挂在天花板上的艺术品。您或许还记得自己卧室天花板上挂着的一个。

A mobile consists of sticks and figures. Each stick has a few points to which figures or other sticks may be attached.

一个活动挂图由木棍和图形组成。每根木棍上有几个点，图形或别的木棍可以连接在这些点上。

Also, each stick has a hanging point, from which it is attached to a stick further above (or to the ceiling).

此外，每根棍子都有一个悬挂点，通过这个点它被连接到上方更高处的棍子（或者天花板）。

The following example mobile can be described using these numbers and brackets:

以下这个示例手机可以用这些数字和括号来描述：

(-3 (-1 1) (1 1)) (2 3)

Question: 问题:

Which of the following mobiles is described by the following instructions:

以下哪款手机符合以下描述：

(-3 (-1 4) (2 (-1 1) (1 1))) (2 (-1 6) (2 3))

A.

B.

C.

D.

 

 

 

 

 

  Explanation 解释

正确的答案是A.

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（Bebras每日一练部分展示）

Every Friday, six spies exchange all the information they have gathered during the week. A spy can never be seen with more than one other spy at the same time. So, they have to have several rounds of meetings where they meet up in pairs and share all the information they have at that point.

每周五，六名间谍都会交换他们一周内所收集到的所有情报。一名间谍绝不能在同一时间与超过一名间谍碰面。因此，他们得分多轮进行会面，每次两人一组，分享当时各自所掌握的所有情报。

The group of 6 spies needs only three rounds to distribute all secrets:

这 6 名间谍组成的小组仅需三轮就能分发完所有机密：

Before the meetings each spy holds a single piece of information. (spy 1 knows 'a', spy 2 knows 'b, etc.). In the first round spies 1 and 2 meet and exchange information so now both know 'ab'. The diagrams show which spies meet in each round with a line. It also shows which pieces of information they all have. After three rounds all information has been distributed.

在会议开始前，每个间谍都掌握着一条单独的信息。（间谍 1 知道“a”，间谍 2 知道“b”，以此类推）。在第一轮中，间谍 1 和间谍 2 相遇并交换信息，这样他们俩现在都知道“ab”。图表显示了每一轮中哪些间谍会面，用一条线表示。它还展示了他们所有人所拥有的信息。经过三轮，所有信息都已分发完毕。

Question: 问题:

After an international incident one spy has stopped attending the meetings. What is the minimum number of rounds needed for the five remaining spies to exchange all information?

在一次国际事件之后，一名间谍不再参加会议。那么剩下的五名间谍要交换所有信息，最少需要几轮？

 

 

 

 

 

 

Explanation 解释

Correct answer is 4 rounds.

正确答案是 4 轮。

This is unexpected: the obvious answer is three (or less?) since we have one spy less. This is even stranger if we consider that four spies would quite obviously exchange the information in two rounds.

这出乎意料：显而易见的答案是三轮（或者更少？），因为我们少了一个间谍。如果考虑到四个间谍显然只需两轮就能交换完信息，那就更奇怪了。

However, unsuccessful attempts at solving the task soon show us the root of the problem: since the number of spies is odd, one of them is “inactive” in every round. Say that spy number 5 does not participate in the first round, but he participates in the second. Thus in the second round, only two spies will know his piece information (e). In the third round, these two spies will meet two other spies, so after three rounds (only) four spies will know e. The fourth round is needed to spread this information to the fifth.

然而，几次尝试解决该任务均未成功，很快我们就发现了问题的根源：由于间谍的数量是奇数，所以在每一轮中都会有一个间谍处于“不活跃”状态。比如，间谍 5 在第一轮不参与，但在第二轮参与。因此，在第二轮中，只有两个间谍会知晓他的情报（e）。到了第三轮，这两个间谍会分别与另外两个间谍碰面，所以经过三轮（仅）之后，只有四个间谍知晓 e。需要第四轮才能将此信息传递给第五个间谍。

Therefore, we proved that at least four rounds are needed. To show that they also suffice, we construct a schema with four rounds.

因此，我们证明了至少需要四轮。为了表明四轮也足够，我们构建了一个四轮的模式。

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（Bebras每日一练部分展示）