分类： 赛事动态

A robot can move around a 10 x 10 gray grid and paint black or white squares.一个机器人可以在一个 10×10 的灰色网格上移动，并将方格涂成黑色或白色。

The robot can only move up, down, left, or right, and one square at a time.这个机器人只能向上、向下、向左或向右移动，每次只能移动一格。

An experiment to produce some interesting patterns was performed:进行了一项实验以生成一些有趣的图案：

• Some black dots and white dots are placed randomly on the grid.一些黑色的点和白色的点被随意地放置在网格上。
• The robot is programmed to move from left to right across the top row, then right to left across the second row, then left to right across the third row, and so on. This way it snakes across all 100 cells of the grid.这个机器人被设定为先从左向右移动穿过最上面一行，然后从右向左移动穿过第二行，接着从左向右移动穿过第三行，依此类推。这样它就能蜿蜒穿过整个网格的 100 个单元格。
• Different programs are run to see what patterns can be made by the robot.运行不同的程序，看看机器人能做出什么样的图案。

Programs: 程序:

A. The cells are painted the color of the nearest dot to the robot. If there is a tie, no color is painted.A. 单元格会被涂成离机器人最近的点的颜色。如果距离相同，则不涂色。

B. The cells are painted the color of the furthest dot from the robot. If there is a tie, no color is painted.B. 单元格会被涂成离机器人最远的那个点的颜色。如果出现平局，则不涂色。

C. The cells are painted the color of the last dot it passes over.C. 单元格会被涂成它最后经过的那个点的颜色。

D. The robot looks at all the cells it could go to in two moves or less, from its current position. It paints this cell the color of the most common color dot it "sees" in these cells. If there is a tie, or there are no dots in these cells, no color is painted.D. 机器人从其当前所在位置查看两步或两步以内能够到达的所有单元格。它将这些单元格涂成在这些单元格中“看到”的最常见颜色的点的颜色。如果出现平局，或者这些单元格中没有点，则不涂色。

Task:任务:

Given the positions of the randomly placed dots, shown above, match the patterns to the programs the robot followed.根据上图中随机放置的点的位置，将图案与机器人遵循的程序相匹配。

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

 

 

 

 

 

 

Explanation 解释

2022-2023年Bebras挑战练习题

Juniors (14-16岁) 集合

名称	类型
Cipher 8 密码8	A
Lights On 亮着灯	A
Flower Garden 花园	A
Nuts and Bolts 螺母和螺栓	A
Mary's Neighbors 玛丽的邻居	A
Listen and Walk 听和走	B
Tic-Tac-Toe 井字	B
Hangar Carousel 机库旋转木马	B
Lists 列表	B
Favorite Movie 最喜欢的电影	B
Favorite Gem 最喜欢的宝石	C
Maze 迷宫	C
Mysteria	C
Treasure Box 宝箱	C
Packing 包装	C

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A rectangular gift box is needed to hold 4 different sized chocolate bars. The biggest bar has 15 pieces.需要一个长方形的礼品盒来装 4 根不同尺寸的巧克力棒。最大的那根巧克力棒有 15 块。   Chocolate bars cannot be stacked on top of each other, and the gift box should have as few gaps as possible.巧克力棒不能叠放在一起，而且礼品盒内的空隙应尽可能少。   For example, if the chocolate bars are arranged as shown on the right, there are 7 gaps.例如，如果巧克力棒像右边那样排列，就有 7 个空隙。

Task 任务:

Arrange the chocolate bars in a rectangle, with as few gaps as possible.把巧克力棒摆成一个长方形，尽量少留空隙。

(Drag the chocolate bars by the '+'. Rotate them by clicking the ↻. Press 'Save' when you are finished.)（通过“ ”拖动巧克力棒。点击“↻”旋转它们。完成后点击“保存”。）

 

 

 

 

 

 

Explanation 解释

Answer:答案:

The best possible arrangements leave only 2 gaps.最好的安排也只能留出两个空位。

Explanation:解释:

One possible arrangement:一种可能的安排：

This leaves us with 2 gaps in the top left of the rectangle.这在矩形的左上角给我们留下了两个空缺。

The number of chocolates pieces in all 4 bars put together is 12 + 15 + 6 + 5 = 38. A gift box that holds 38 pieces with 0 gaps must have dimensions 1 x 38 or 2 x 19. You will never be able to fit the 3 x 5 chocolate bars (or the 3 x 4 chocolate bar) in this gift box.将这 4 块巧克力合在一起，巧克力块的总数为 12 + 15 + 6 + 5 = 38 块。一个能容纳 38 块巧克力且没有空隙的礼盒，其尺寸必须是 1×38 或 2×19。你永远无法将 3×5 的巧克力块（或 3×4 的巧克力块）装进这个礼盒。

A gift box with 1 gap would hold 39 chocolate pieces. There are two possibilities, 1 x 39 (not possible) or 3 x 13. The two largest chocolate bars would take up 9 rows within such a gift box. The remaining 4 rows are not enough to place the smallest chocolate bar of size 1 x 5.一个有 1 个空隙的礼盒能装 39 块巧克力。有两种可能的排列方式，1×39（不可能）或 3×13。礼盒中最大的两块巧克力会占据 9 行。剩下的 4 行不足以放置最小的 1×5 的巧克力。

Hence, 2 gaps is the minimum we can have in any gift box, and we can achieve this as shown above.因此，任何礼品盒中至少会有 2 个空隙，而且如上图所示，我们能够达到这个最小值。

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Maria found a box containing hidden treasure, but the box was locked.玛丽亚发现了一个装有隐藏宝藏的盒子，但盒子是锁着的。

To unlock the box, she needs to use the correct combination of three shapes.要打开这个盒子，她需要使用三个图形的正确组合。

Help Maria unlock the box by following the given hints on the right of the combinations shown below.请按照下面组合密码右侧给出的提示帮助玛丽亚打开这个盒子。

Question:问题:

Which of the following combinations will unlock the treasure box?以下哪一种组合能打开宝箱？

A.

B.

C.

D.

   

 

 

 

 

 

Explanation 解释

Answer: 答案:

Explanation: 解释:

We will start by eliminating the shapes that do not fit into the combination that unlocks the box.我们将从排除那些不符合能打开盒子的组合的形状开始。

In the second row of hints, we see that nothing belongs in our combination, which means that the Christmas three , diamond  and arrow  do not lead to unlocking the box.在第二行提示中，我们看到没有任何东西属于我们的组合，这意味着圣诞树、菱形和箭头都不能打开这个盒子。

In the last row of hints, we see that one shape is correct, but placed incorrectly.在最后一行提示中，我们看到有一个形状是正确的，但位置放错了。

We have previously concluded that the Christmas tree and the arrow are not used, so the shape we need is a star , but it is placed incorrectly.我们之前已经得出结论，圣诞树和箭头没有用到，所以我们需要的形状是星星，但它的位置放错了。

The possible positions of the start are as follows:起点可能的位置如下：

 or 或We can continue to find the other two shapes.我们可以继续找出另外两种形状。

From the first row, we see that one form is correct and in the right place. We reject the arrow because it does not belong to the final combination, so it follows that one of these combinations is correct:从第一行来看，有一种形式是正确的，并且处于正确的位置。我们排除箭头，因为它不属于最终的组合，所以可以得出结论，这些组合中有一个是正确的：

 or 或From the third row, we see that two shapes are correct, but placed incorrectly.从第三行来看，有两个图形是正确的，但位置放错了。

We definitely need the star, but its position is not in the middle, so now we have definitely found the position of the star and that position is given in the following picture:我们确实需要那颗星，但它的位置不在中间，所以现在我们已经确切地找到了那颗星的位置，其位置在下面的图片中给出：

We found one shape and it's correct position. Likewise, we continue this process to find the remaining two shapes.我们找到了一个形状及其正确位置。同样地，我们继续这个过程来寻找剩下的两个形状。

From the first row, we had:从第一行，我们有： or 或

Since we found the star to be in the first position, the triangle can't be in the first position. So, it follows that the moon is in its correct position. This leaves us with:既然我们发现星星处于第一个位置，那么三角形就不能在第一个位置。所以，月亮就在其正确的位置。这样就剩下：

From the fourth hint, we see that one shape is correct, but incorrectly placed. The triangle  is out, and the only position left is the middle position. The heart  can't be correct, because it is in the middle position. Thus, the only shape that can be in the middle position is the circle .从第四个提示中可以看出，有一个图形是正确的，但位置放错了。三角形被排除在外，剩下的唯一位置就是中间位置。心形不可能正确，因为它在中间位置。所以，唯一能放在中间位置的图形就是圆形。

There are other possible ways to determine the answer other than the explanation above. All these different ways will still lead to the correct answer.除了上述解释之外，还有其他可能的方法来确定答案。所有这些不同的方法最终都会得出正确的答案。

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（Bebras每日一练部分展示）

In a land called Mysteria, there lives a wizard. 在一片名为神秘之地的大陆上，住着一位巫师。

The wizard can transform items into an assortment of other items. The order of these items will always be the same.这位巫师可以把物品变成各种各样的其他物品。这些物品的排列顺序总是固定的。

For example: if a coin is transformed into a potion and a dragon, the potion will always be on the left of the dragon.例如：如果一枚硬币被转化成一瓶药水和一条龙，那么药水总是会在龙的左边。

The table below shows the two possible ways a wizard can transform a diamond and the two possible ways he can transform a coin:下表展示了巫师将钻石进行转化的两种可能方式以及将硬币进行转化的两种可能方式：

Before 之前	After 后

  These magical transformations can happen any number of times, in any order. That is, any diamond and any coin can be transformed at any time. In this way, a long line of magical things can be created from a single diamond or coin.这些神奇的变化可以以任意次数、任意顺序发生。也就是说，任何一颗钻石和任何一枚硬币都可以在任何时候进行变化。通过这种方式，从一颗钻石或一枚硬币出发，可以创造出一长串神奇的事物。

Question:
问题:

Starting with a single diamond, which line is not possible to obtain?从一颗钻石开始，哪一行是无法得到的？

A.

B.

C.

D.

   

 

 

 

 

 

Explanation 解释

The correct answer is: 正确答案是：

Suppose the magical transformations are numbered 1 through 4 as follows:假设这些神奇的变形按如下方式编号为 1 至 4：

Number 数量	Before 之前	After 后
1
2
3
4

can be obtained by starting with a single wizard and applying transformations, 1, 4, 2 and 3 in that order.可以通过从一个巫师开始，并依次应用变换 1、4、2 和 3 来获得。

can be obtained by starting with a single wizard and applying transformations 2, 2, 3, 4, and 1 in that order.可以通过从一个巫师开始，并依次应用变换 2、2、3、4 和 1 来获得。

can be obtained by starting with the single wizard and applying transformations 2, 1, 3 and 3 in that order.可以通过从单个巫师开始，并依次应用变换 2、1、3 和 3 来获得。

One quick way to see that  is not possible is to notice that the transformation rules always create a potion and a dragon at the same time. Therefore, the number of potions in Mysteria will always equal the number of dragons, which is not the case in this answer option.要快速看出这是不可能的，只需注意到转化规则总是同时生成一瓶药剂和一条龙。因此，神秘岛上的药剂数量总是等于龙的数量，而这在该选项中并非如此。

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The Little Beaver is in a maze. The maze is made up of two floors, each with its own grid of obstacles.小海狸被困在了一个迷宫里。这个迷宫由两层组成，每层都有自己的障碍网格。

The Little Beaver can move between two adjacent cells within one floor if there is no wall between the cells; this takes one second.小海狸可以在同一层内相邻的两个格子之间移动，前提是这两个格子之间没有墙；这需要花费一秒钟。

The Little Beaver can also use her magic wand to move to the corresponding cell of the other floor; this takes five seconds.小海狸还可以用她的魔法棒移动到另一层对应的单元格；这需要五秒钟。

For example, if the Little Beaver is in cell A, there are three possible moves:例如，如果小海狸在 A 单元格，那么有三种可能的移动方式：

1. Move left. This move takes 1 second.1. 向左移动。此动作耗时 1 秒。

2. Move down. This move takes 1 second.2. 向下移动。此动作耗时 1 秒。

3. Move to the corresponding cell of the other floor. This move takes 5 seconds.3. 移动到另一层对应的单元格。此移动耗时 5 秒。

The Little Beaver starts at cell A and wants to reach cell B as soon as possible. 小海狸从 A 单元格出发，想要尽快到达 B 单元格。

Question:问题:

What is the shortest time needed for the Little Beaver to reach cell B, if starting from cell A?如果小海狸从 A 单元格出发，那么它到达 B 单元格所需的最短时间是多少？

A. 16

B. 17

C. 18

D. 20

 

 

 

 

 

Explanation 解释

Troy has a collection of gems.特洛伊收藏了一些宝石。

He ranks his gems from his overall favorite to his least favorite.他将自己收藏的宝石从最喜爱的到最不喜爱的依次排列。

Sarah knows which gems are in Troy's collection, but she does not know how he has ranked them.萨拉知道特洛伊收藏的宝石都有哪些，但她不知道他是如何给这些宝石排序的。

Sarah has a plan to find out which gem is Troy's overall favorite:萨拉有个计划，要弄清楚特洛伊最喜欢哪种宝石：

• Sarah chooses four of Troy's gems and asks Troy: "Out of this group of four, which gem is your favorite?"萨拉从特洛伊的宝石中挑了四颗，然后问特洛伊：“在这四颗宝石中，哪一颗是你最喜欢的？”
• Sarah chooses a new set of four gems and asks her question again.萨拉挑选了一套新的四颗宝石，再次提出了她的问题。
• Then she chooses a third set of four gems and asks her question for the last time.然后她挑选出第三组四颗宝石，并最后一次提出她的问题。

Note: When Sarah chooses her second and third set of four gems, she may sometimes include gems she has chosen before.注意：当萨拉选择她的第二组和第三组四颗宝石时，有时可能会包含之前选过的宝石。

Question:问题:

If Sarah is to successfully find Troy's overall favorite gem, what is the largest possible number of gems in Troy's collection?如果萨拉要成功找到特洛伊最喜欢的宝石，那么特洛伊收藏的宝石最多可能有多少颗？

A. 8

B. 10

C. 11

D. 12

 

 

 

 

 

Explanation 解释

Answer: 答案:

10

Explanation: 解释:

With 10 gems, Sarah can ask Troy about eight different gems with the first two requests. Troy's answer to each of these requests is a candidate to be his favourite of all his gems, but the other three cannot be. Therefore, with Sarah's third and final request, she can include these two candidates and the two gems that have not yet been part of a request. Troy's answer to this third request must be his favourite gem.萨拉有 10 颗宝石，她可以用前两次询问来询问其中的 8 颗。特洛伊对这两次询问的回答可能是他所有宝石中最喜欢的那颗，但另外三颗则不可能。因此，在萨拉的第三次也是最后一次询问中，她可以把这两个可能的答案以及之前未被询问过的两颗宝石都包括进去。特洛伊对这次询问的回答一定是他最喜欢的那颗宝石。

We have shown that there is a strategy of Sarah's that works if there are 10 gems. (Note that there are other correct strategies that work for 10 gems.)我们已经证明，如果宝石有 10 颗，那么萨拉有一种策略是可行的。（请注意，对于 10 颗宝石，还有其他正确的策略也是可行的。）

If Troy has at least 11 gems, we consider Sarah's first two requests.如果特洛伊至少有 11 颗宝石，我们就考虑萨拉的前两个请求。

If at least one gem is part of both requests, then at least four gems will be unconsidered after the first two requests. In this case, Sarah must ask about these four unconsidered gems, because if she doesn't, an unconsidered gem could be Troy's favourite. On the other hand, she won't have any information about Troy's favourite among the last four unconsidered gems compared to the other seven gems. In this case, her strategy cannot determine which gem is Troy's most favourite.如果至少有一颗宝石同时属于两个请求，那么在前两个请求之后，至少会有四颗宝石未被考虑。在这种情况下，萨拉必须询问这四颗未被考虑的宝石，因为如果不这样做，未被考虑的宝石中可能就有特洛伊最喜欢的那颗。另一方面，与另外七颗宝石相比，她对最后四颗未被考虑的宝石中哪一颗是特洛伊最喜欢的毫无头绪。在这种情况下，她的策略无法确定哪颗宝石是特洛伊最喜欢的。

If there is no gem that is part of Sarah's both first two requests, then candidates for Troy's favourite include the two answers to these requests and the three remaining unconsidered gems. This is a total of five possibilities for Troy's most favourite gem, and Sarah does not have any information about how they rank, so her strategy cannot work.如果莎拉最初的两个请求中没有共同包含的宝石，那么特洛伊最喜欢的宝石候选者就包括这两个请求的答案以及剩下的三颗未被考虑的宝石。这样特洛伊最喜欢的宝石就有五种可能性，而莎拉对它们的排名毫无头绪，所以她的策略就行不通了。

We have shown that there is not a strategy of Sarah's that works, if there are more than 10 gems.我们已经证明，如果宝石数量超过 10 颗，那么萨拉的任何策略都无法奏效。

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（Bebras每日一练部分展示）

2022-2023年Bebras挑战练习题

Cadets (12-14岁) 集合

名称	类型
Hamburger Recipe 汉堡包配方	A
Flower Garden 花园	A
Sailor Necklace 水手项链	A
Embroidery Machine 绣花机	A
Pantry Map 储藏室地图	A
Rug Weaving 地毯纺织	B
Nuts and Bolts 螺母和螺栓	B
Lights On 亮着灯	B
Cipher 8 密码8	B
Mary's Neighbors 玛丽的邻居	B
Lists 列表	C
Listen and Walk 听和走	C
Tic-Tac-Toe 井字	C
Favorite Movie 最喜欢的电影	C
Hangar Carousel 机库旋转木马	C

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（Bebras每日一练部分展示）

At Beavertown Airfield, six planes are parked on a rotating turntable in a round hangar.在比弗敦机场，六架飞机停放在圆形机库中一个可旋转的转盘上。

The turntable can be rotated left or right by using a control panel with two arrows ◄►. One button press rotates the turntable exactly one parking position, either left or right. The gate of the hangar is wide enough for one plane to roll out. The turntable is very slow to rotate, so having fewer button presses will avoid delays.转盘可通过带有两个箭头“◄►”的控制面板向左或向右旋转。按一次按钮，转盘就会精确地旋转一个停车位的位置，要么向左，要么向右。机库的门足够宽，可供一架飞机驶出。转盘旋转速度非常慢，因此减少按钮按压次数可避免延误。

In the mornings, when pilots come to pick up their planes, the parking position 1 is always at the gate. In the best case scenario, the arrow keys need to be pressed five times to get all planes to roll out. In these cases pilots will access the parking positions in the order: 1, 2, 3, 4, 5, 6 by pressing ► five times, or in the order: 1, 6, 5, 4, 3, 2 by pressing ◄ five times.早上，当飞行员来取飞机时，1 号停机位总是在登机口。在最佳情况下，需要按五次方向键才能让所有飞机驶出。在这种情况下，飞行员会按顺序访问停机位：1、2、3、4、5、6，方法是按五次“右箭头”键，或者按顺序访问停机位：1、6、5、4、3、2，方法是按五次“左箭头”键。

But what is the worst case scenario? That is, what order of accessing the parking positions will require the maximum number of button presses for all planes to be rolled out?但最糟糕的情况会是怎样？也就是说，哪种取车顺序会让所有飞机都驶出车位所需的按键次数最多？

Task:任务:

Provide one of such worst case order for pilots to access the parking positions 1 - 6 by dragging the numbers into the right spots.请将飞行员进入 1 至 6 号停机位的最糟糕的顺序之一通过将数字拖到正确的位置来提供。

 

 

 

 

 

Explanation 解释

A group of friends wants to choose which of seven movies to watch.一群朋友想要从七部电影中选一部来看。

Each friend rates each movie from highest to lowest: , ,  .每位朋友都按照从高到低的顺序给每部电影打分。

A movie is called a "favorite movie" if none of the friends has given another movie a higher rating.如果朋友中没有人给另一部电影打出更高的评分，那么这部电影就被称为“最喜爱的电影”。

Example:例子:

Movie 1 is NOT a favorite movie because Niklaus gave a higher rating to Movie 4.电影 1 不是最喜欢的电影，因为尼克劳斯给电影 4 的评分更高。

Task:任务:

Make as few changes as possible to the ratings so there is a favorite movie.尽量少改动评分，以便能选出一部最受欢迎的电影。

There are multiple ways to do this.有多种方法可以做到这一点。

(Click on a rating to change it and press 'Save' when you are finished.)（点击一个评分来更改它，完成后点击“保存”。）

 

 

 

 

 

Explanation 解释