分类： 赛事动态

There are 10 plates in a row. There is one apple on each plate.

一排有 10 个盘子。每个盘子上都有一个苹果。

Thomas, the kangaroo, loves to jump. First he jumps onto the leftmost plate with the letter A.

袋鼠托马斯喜欢跳跃。首先，它跳到了最左边标有字母 A 的盘子上。

On each single jump after this, he either jumps forward two plates, or backwards three plates.

在此之后的每一次跳跃，他要么向前跳两块板，要么向后跳三块板。

(An example of the two possible jumps from one plate is shown with arrows in the picture.)（从一块板上可能做出的两种跳跃方式在图中用箭头标出。）

Thomas only jumps onto plates with an apple. If he jumps onto a plate, he collects the apple from it.

托马斯只跳到有苹果的盘子上。如果他跳到盘子上，就会从盘子里把苹果拿走。

Question:问题:

If Thomas collects all 10 apples, which apple does he collect last?

如果托马斯把 10 个苹果都收集起来，他最后收集的是第几个苹果？

 

 

 

 

 

 

Explanation 解释

The correct answer is: I (the second apple fromt he right)

正确答案是：I（从右数第二个苹果）

We number the plates from left to right from A to J. Then Thomas can collect all ten apples in the order A, B, C, D, E, F, G, H, I, J.

我们从左到右将盘子依次编号为 A 到 J。这样托马斯就可以按照 A、B、C、D、E、F、G、H、I、J 的顺序收集所有十个苹果。

This is the only sequence of jumps that allows Thomas to collect all the apples. Why? To begin, Thomas must jump on plates A, C and then E because otherwise he jumps to the left of the first plate. Next, he must jump to plate B because he can only get to plate B from plate E and he will not return to plate E later. The same kind of reasoning can be used to see that all the remaining jumps are also determined uniquely.

这是托马斯能收集到所有苹果的唯一跳跃顺序。为什么？首先，托马斯必须跳到 A、C 然后是 E 板上，否则他会跳到第一块板子的左边。接下来，他必须跳到 B 板上，因为他只能从 E 板跳到 B 板，而且之后不会再回到 E 板。同样的推理可以得出，所有剩余的跳跃也都唯一确定了。

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（Bebras每日一练部分展示）

Beaver the Alchemist can convert objects into new objects. He can convert:

炼金术士海狸能够将物品转换成新的物品。他可以转换：

• Two clovers into a coin两片三叶草变成一枚硬币
• A coin and two clovers into a ruby一枚硬币和两片三叶草变成了一颗红宝石
• A ruby and a clover into a crown将一颗红宝石和一株三叶草制成一顶王冠
• A coin, a ruby, and a crown into a kitten.一枚硬币、一颗红宝石和一顶王冠变成了一只小猫。

After an object has been converted into another object, it disappears immediately.一个物体在被转换成另一个物体之后，会立刻消失。

Question: 问题:

How many clovers does Beaver the Alchemist need to create one kitten?炼金术士海狸要创造一只小猫需要多少株三叶草？

A. 5

B. 10

C. 11

D. 12

 

 

 

 

 

Explanation 解释

The answer is 11. We can see the conversion as follows:

答案是 11。我们可以看到转换过程如下：

coin       =   2 clovers

ruby      =   2 clovers + 1 coin = 4 clovers

crown    =   1 ruby + 1 clover = 4 clovers + 1 clovers = 5 clovers

kitten    =   1 coin + 1 ruby + 1 crown = 2 clovers + 4 clovers + 5 clovers = 11 clovers

The answer 5 is for those who did not read the graph properly to realize that we need 2 clovers to create a coin or ruby, or for those who did not realize that you need a coin to create a ruby or a ruby to create a crown.

答案 5 是给那些没有正确读图从而没有意识到需要 2 片三叶草才能合成一枚硬币或一颗红宝石的人，或者是那些没有意识到需要一枚硬币才能合成一颗红宝石，或者需要一颗红宝石才能合成一顶王冠的人。

The answer 10 is for those who think that the number of clovers is the same as the number of edges.

答案 10 是给那些认为三叶草的数量与边的数量相同的那些人的。

The answer 12 is for those who miscounted.

答案 12 是给那些数错的人的。

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（Bebras每日一练部分展示）

The teacher at the beaver school wants to give some material to his students.

海狸学校的老师想给他的学生们一些材料。

He found a portal with a scanned book that explains in its front page that it should be distributed according to a “Creative Commons License” (CC-BY-ND). This allows everyone to share, copy and redistribute the material for free  in any medium or format for any purpose, even commercially, provided that appropriate credit is given.

他发现了一个带有扫描书籍的入口，该书的首页说明应根据“知识共享许可协议”（CC-BY-ND）进行分发。这允许任何人免费在任何媒介或格式中分享、复制和重新分发该材料，用于任何目的，甚至商业用途，但需给予适当的署名。

The license also specifies that if one remixes, translates, or builds upon the book, the modified book may not be distributed.

该许可证还规定，如果有人对这本书进行改编、翻译或在其基础上进行创作，改编后的作品不得进行分发。

Question: 问题:

Which of these actions is not permitted under the terms of this license?

根据本许可条款，以下这些行为中哪一项是不被允许的？

A. Selling book copies to the students.

B. Translating the book, keeping the translated copy for himself.

C. Giving the students his translation of one chapter of the book.

D. Putting a copy of the scanned book on the school website.

 

 

 

 

 

Explanation 解释

C is the correct answer.C 是正确答案。

2015年Bebras挑战练习题

Junior (14-16岁) 集合

名称	类型
Irrigation System（灌溉系统）	A
Fair share （公平的分享）	A
Super Power Family （超能力家族）	A
E-mail Scam （电子邮件诈骗）	A
Beaver Logs （海狸日志）	A
Theatre （剧院）	B
Beaver Lunch （海狸的午餐）	B
Stack Computer （堆栈的电脑）	B
Throw the dice （掷骰子）	B
Drawing Stars （画星星）	B
Popularity （流行）	C
Bowl Factory （碗的工厂）	C
Word Chains （文字链）	C
Hard Fireworks （艰难的烟花）	C
Pirate Hunters （海盗猎人）	C

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（Bebras每日一练部分展示）

In the game of Pirate Hunters players take turns moving a Pirate or a Policeman.在“海盗猎人”游戏中，玩家轮流移动海盗或警察。

When it is the police's turn, the player moves a policeman over to a neighbouring field.轮到警察时，玩家将一名警察移至相邻的区域。

The pirate is faster than a policeman, and skips a field on its turn, moving two fields.海盗比警察快，在其回合时能跳过一格，移动两格。

A policeman cannot move to a field that is occupied -- either by his colleague policeman, or the pirate.一名警察不能移动到已被占据的区域——无论是被他的同事警察占据，还是被海盗占据。

The game ends when the pirate is forced to move to a field occupied by one of the policemen.当海盗被迫移动到一名警察所在的场地时，游戏结束。

The policeman goes first.警察走在前面。

Question: 问题:

If the pirate plays the best way possible and makes no mistakes, how many moves will it take the police to capture him?如果海盗以最佳方式行动且不犯任何错误，警察需要多少步才能将其抓获？

A. The police can win in 2 turns.

B. The police can win in 3 turns.

C. The police can win in 5 turns.

D. The police has no chance of winning.

 

 

 

 

 

Explanation 解释

The answer is: The police cannot win the game if the pirate plays perfectly.

答案是：如果海盗玩得完美无缺，警察就赢不了这场游戏（选项D）。

Let us assume that the police actually forced the pirate into the above situation and it is the pirate's turn (so he loses). What did the board look like before the police's last move? The police moved one or the other policeman up or down. Since the board is symmetric, we shall assume that it was the right policeman that moved; the situation for the left one is the same. Thus the previous situation was one of these.

我们假设是警方迫使海盗陷入上述局面，且轮到海盗走（所以海盗输了）。警方的上一步棋之前棋盘是什么样子的？警方移动了其中一名警察向上或向下。由于棋盘是对称的，我们假设移动的是右边的警察；左边警察的情况是一样的。因此，之前的局面是这样的。Let us go one more move back. The pirate must had come from the right. Hence the situation before that was one of these two:

让我们再回溯一步。海盗必定是从右边来的。所以之前的情况必然是这两种之一：The above situation, in which the pirate is captured, can therefore arise only from these two positions (and their mirrored versions). However, if the pirate is indeed a good player and found himself in one of these two positions, he would surely move the pirate to the left and not up.

上述海盗被擒获的情况因此只能从这两个位置（及其镜像位置）产生。然而，如果海盗确实是个高手，发现自己处于这两个位置中的任何一个，他肯定会把海盗向左移动而不是向上移动。

Two beavers live in lodges separated by a large forest.

两只海狸分别住在被大片森林隔开的窝里。

They decide to send messages to each other by shooting fireworks into the sky above the trees.

他们决定通过向树梢上方的天空发射烟花来互相传递信息。

Each message is a sequence of words, though the beavers only know five different words.

每条信息都是一串单词，不过海狸们只认识五个不同的单词。

The beavers can shoot two types of fireworks, one after the other, and know the following codes:

海狸们能够依次发射两种烟花，并且知晓以下代码：

For example, to send the (rather strange) message "food, log, food", a beaver would shoot:

例如，要发送（相当奇怪的）信息“食物，原木，食物”，一只海狸会这样射击：

Question: 问题:

How many different meanings can the following sequence of fireworks have?

下面这一系列烟花能有多少种不同的含义？

 

 

 

 

 

 

Explanation 解释

The correct answer is 4. The message could mean any of the following:

正确答案是 4。该消息可能表示以下任何一种意思：

• log, rock, food, river 日志，岩石，食物，河流
• log, log, log, river 木头，木头，木头，河流
• rock, tree, river 岩石、树木、河流
• rock,food, log, river 岩石、食物、原木、河流

To convince yourself that there are no more possibilities, you can systematically count them:

要让自己确信没有其他可能性，你可以系统地进行计数：

- Start with the first firework. It is not a message, so you can assign a zero to it.

从第一个烟花开始。它不是信息，所以你可以给它赋值为零。

- The first two fireworks can only mean log. Assign number one to the second firework.

前两束烟花只能表示“木”。给第二束烟花标上数字 1。

- We are now at the third firework. It can have a meaning of any shorter sub-sequence plus one new word. Yet we see that there is no way to prolong the previously examined sequences (of length 1 and 2), so we only have one possible meaning (rock) and assign 1 to the third firework.

现在我们来到了第三个烟花。它可以表示任何更短的子序列加上一个新词的意思。然而我们发现无法延长之前考察过的序列（长度为 1 和 2 的），所以我们只有一个可能的意思（“岩石”），于是给第三个烟花赋值 1 。

- The fourth firework is somewhat interesting! It can either add the word log to the first two fireworks, or food to the first three fireworks, as shown by the arrows below. So we sum the two numbers at the 2^nd and 3^rd firework and assign it to the 4^th (1+1=2).

第四个烟花有点意思！它要么把单词“log”加到前两个烟花上，要么把“food”加到前三个烟花上，如下方箭头所示。所以我们把 2 号和 3 号烟花上的两个数字相加，并把这个和赋值给 4 号烟花（1 + 1 = 2）。

- We proceed applying the same idea to each firework to the right. We look one, two and three fireworks back. If those shorter messages can be prolonged with a correct word, we mark this fact with an arrow. Then we just sum the numbers “brought” by the arrows to the currently examined firework.

我们接着将同样的思路应用于右侧的每一个烟花。我们向后看一个、两个和三个烟花。如果那些较短的信息能用一个正确的词加以延长，我们就用箭头标出这一事实。然后，我们只需将箭头“带来”的数字加到当前正在考察的烟花上。

- At the last firework we will have the number of all possible meanings.

在最后一束烟花绽放时，我们将拥有所有可能意义的数量。

The systematic approach when we build our solution systematically step by step and using the previous steps is called dynamic programming. It makes the process much easier.

当我们逐步系统地构建解决方案，并利用之前的步骤时，这种系统的方法被称为动态规划。它使整个过程变得容易得多。

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（Bebras每日一练部分展示）

For his homework, Thomas had to write words on cards and connect them with rubber bands.托马斯的家庭作业是把单词写在卡片上，然后用橡皮筋把它们连起来。   The teacher told him to connect any two words that differ by exactly one letter.老师让他把仅相差一个字母的任意两个单词连接起来。 Thomas did this, as you can see in the picture on the right.托马斯就是这样做的，如您在右边的图片中所见。 When Thomas returned from having a break he got a surprise.托马斯休息回来时吃了一惊。 Peter, his little brother, had erased all the words!彼得,他的小弟弟把所有的字都擦掉了！
	Also, the cards were completely mixed up, as you can see in the image on the left.而且，正如您在左边的图片中所看到的那样，这些卡片完全被打乱了。 Importantly, the rubber bands still connected them as before.重要的是，橡皮筋仍像之前那样把它们连在一起。 Thomas was sure he could put the words back in the correct place.托马斯确信他能把那些词放回正确的位置。

Question: 问题:

Which of the pictures below contains the words in exactly the right places?

下面哪张图片中的单词位置完全正确？

A.

B.

C.

D.

 

 

 

 

 

Explanation 解释

A factory produces sets of 6 bowls of different sizes. A long conveyor belt moves the bowls one by one, from left to right.

一家工厂生产每套 6 个不同尺寸的碗。一条长长的传送带将碗一个接一个地从左向右移动。

Bowl production places the 6 bowls of each set onto the conveyor belt in a random order.

碗的生产过程是将每套的 6 个碗以随机的顺序放置在传送带上。

Before packing the bowls, they need to be sorted to look like this:

在打包碗之前，需要将它们分类整理成如下这样：

To help with the sorting, the factory places workers along the conveyor belt.

为了便于分类，工厂在传送带上安排了工人。

When a set of bowls passes a worker, he will swap any two neighboring bowls which are in the wrong order.

当一组碗经过一名工人时，他会交换任何两个顺序错误的相邻碗。

He will keep doing this until the set of 6 bowls has finished passing him.

他会一直这样做，直到那套 6 个碗全部从他面前传过。

See how the order of a set of bowls changes as it passes one worker: 

观察一组碗在经过一名工人时其顺序是如何变化的：

Question: 问题:

How many workers should be put along the line to sort the following set of bowls?

要将下面这组碗进行分类，应该在流水线上安排多少名工人？

 

 

 

 

 

Explanation 解释

The answer is: 4 

答案是：4

As shown in the question, the original order of the set of bowls is: 5 6 3 2 1 4

如问题中所示，碗的原始排列顺序为：5 6 3 2 1 4

Remember that swapping of neighboring bowls happens from right to left.

记住，相邻碗的交换是从右往左进行的。

After passing a first worker, the order of the bowls is: 1 5 6 3 2 4 (4 swaps, all with bowl 1)

经过第一个工人处理后，碗的顺序是：1 5 6 3 2 4（共交换了 4 次，都是与 1 号碗交换）

After passing a second worker, the order is: 1 2 5 6 3 4 (3 swaps, all with bowl 2)

经过第二个工人之后，顺序为：1 2 5 6 3 4（共交换了 3 次，都是与 2 号碗交换）

After passing a third worker: 1 2 3 5 6 4 (2 swaps, all with bowl 3)

经过第三个工人之后：1 2 3 5 6 4（共进行了 2 次交换，均与第 3 个工人交换）

After passing a fourth worker, the set of bowls is sorted: 1 2 3 4 5 6 (2 swaps, all with bowl 4)

经过第四位工人后，碗的排列顺序已整理好：1 2 3 4 5 6（共交换了两次，都是与 4 号碗交换）

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（Bebras每日一练部分展示）

Seven beavers are in an online social network called Instadam.

七只海狸在一个名为“Instadam”的在线社交网络里。

Instadam only allows them to see the photos on their own page and on their friends' pages.Instadam

只允许他们查看自己页面和朋友页面上的照片。

In this diagram, if two beavers are friends they are joined by a line.

在这个图中，如果两只海狸是朋友，它们之间就会有一条线相连。

After the summer holidays everybody posts a picture of themselves on all of their friends' pages.

暑假过后，每个人都会在所有朋友的主页上发一张自己的照片。

Question: 问题:

Which beaver's picture will be seen the most?

哪只海狸的照片会被看到最多？

A. Ari

B. Bob

C. Chio

D. Dmitri

E. Ehab

F. Fritz

G. Gerald

 

 

 

 

 

Explanation 解释

The correct answer is Chio.

正确答案是 Chio。

In order to find the beaver whose picture gets seen by most beavers, you have to count the beavers that are at most two steps away. The beavers that are one step away are those on whose page the pictures will be posted and the beavers that are two steps away are those who can see these pages.

为了找出其照片被最多海狸看到的那只海狸，你得计算出距离它最多两步的所有海狸。距离为一步的海狸是指那些照片会发布在其页面上的海狸，而距离为两步的海狸则是那些能看到这些页面的海狸。

当然，任何一只海狸只能算一次。

The following table shows the names of these beavers as from the starting point of each beaver.

下表列出了从每只海狸的起点算起这些海狸的名字。

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（Bebras每日一练部分展示）

Stella the beaver loves to draw stars. She has devised a system for labeling her stars according to their shape. She uses two numbers:

海狸斯特拉喜欢画星星。她设计了一套根据星星形状给它们编号的系统。她使用两个数字：

• A number of dots for the star.给这颗星星画上一些点。
• A number indicating if a line from a dot is drawn to the nearest dot (the number is 1), the second closest dot (the number is 2), etc.一个数字，表示从一个点到最近的点（数字为 1）、第二近的点（数字为 2）等所画的线。

Here are four examples of Stella's labeling system:    

以下是斯特拉标签系统的四个示例：

Question: 问题:

How would Stella label the following star?

斯特拉会如何给下面这颗星命名？

A. 9:3

B. 9:4

C. 10:4

D. 10:5

 

 

 

 

 

Explanation 解释

The correct answer is:10 : 4 

正确答案是：10 : 4 

The star has 10 dots and it is possible to draw a line to the fourth nearest dot.

这颗星有 10 个点，可以画一条线连到第四个最近的点。

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（Bebras每日一练部分展示）