海狸Bebras国际计算思维挑战赛官网提供Bebras国际计算思维挑战赛的赛事报名通知、详细赛程安排以及相关的培训活动等资讯，为参赛者提供Bebras赛事全面指导。
欢迎使用2015年国际Bebras计算思维挑战赛官方发布的分龄练习题库。本题库按学生年龄划分为五个组别（8-10岁至16-18岁），每组均包含一系列典型题目，可用于：
理解Bebras题型与考察要点；
开展计算思维分阶教学与练习；
评估学生在问题抽象、逻辑推理、模式识别等方面的能力。
您可按对应年龄组选用题目，作为教学、自学或备赛的参考资源。
2015年Bebras挑战练习题-Castors (8-10岁) 集合
2015年Bebras挑战练习题-Benjamin (10-12岁) 集合
2015年Bebras挑战练习题-Cadet (12-14岁) 集合
2015年Bebras挑战练习题-Junior (14-16岁) 集合
2015年Bebras挑战练习题-Senior (16-18岁) 集合
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（Bebras每日一练部分展示）
*练习用* 2022 - 2023 年度Bebras挑战赛
点击一个年龄组即可开始该组别的比赛。
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（Bebras每日一练部分展示）
2022-2023年Bebras挑战练习题
Seniors (16-18岁) 集合
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（Bebras每日一练部分展示）
In the following diagram, the circles are computers and the lines are connections between them.在下面的图表中，圆圈代表计算机，线条代表它们之间的连接。
Computer A is infected with the StuxNet virus and Computer B is infected with the RoXX3 virus.计算机 A 感染了“震网”病毒，计算机 B 感染了“RoXX3”病毒。
At the beginning of each new day, each virus spreads to any computers that are directly connected to an infected computer. If both viruses end up on a single computer, that computer is automatically destroyed and no further spreading will occur from that computer. After a few days, every computer is either infected or destroyed. 在每一天的开始，每种病毒都会传播到与受感染计算机直接相连的任何计算机上。如果两种病毒最终都感染了同一台计算机，那台计算机就会自动被摧毁，且不会再从该计算机传播。几天之后，每台计算机要么被感染，要么被摧毁。
Question:问题:
How many computers are operational but infected with the RoXX3 virus after a week?一周后，有多少台计算机处于可运行状态但感染了 RoXX3 病毒？
You can use the graphic above to find the correct answer by clicking on the circles.您可以点击上方图形中的圆圈来找到正确答案。
(Click on the correct number and your answer will be saved automatically. You can click on another number to change your answer.)（点击正确的数字，您的答案将自动保存。您也可以点击另一个数字来更改答案。）
A. 6
B. 7
C. 8
D. 9
E. 10
Explanation 解释
Answer: 答案:
6 computers are still operating but infected with RoXX3.6 台计算机仍在运行，但已感染了 RoXX3 病毒。
Explanation: 解释:
Day 1: B,C and D are infected with StuxNet and A and E with RoXX3, destroying A and B.第一天：B、C 和 D 感染了“震网”病毒，A 和 E 感染了 RoXX3 病毒，导致 A 和 B 被摧毁。
Day 2: F,G are infected (by C&D respectively) and H & L (by E) with RoXX3, no destructions.第 2 天：F 和 G（分别被 C 和 D 感染）以及 H 和 L（被 E 感染）感染了 RoXX3，没有发生毁灭事件。
Day 3: StuxNet : F and G infect H,I,K (C is already infected).第 3 天：震网病毒：F 和 G 感染 H、I、K（C 已经被感染）。
RoXX3: L infects N and P (E infected already) , H infects N,F.RoXX3：L 感染了 N 和 P（E 已经被感染），H 感染了 N 和 F。
So H and F are destroyed.所以 H 和 F 被摧毁了。
Day 4:StuxNet : K infects J.第 4 天：震网病毒：K 感染了 J。
RoXX3: N infects O,M. RoXX3：N 感染了 O 和 M。
Day 5: StuxNet: J infects M, destroying it.第五天：震网病毒：J 感染了 M 并将其摧毁。
RoXX3: O can infect Q (and N, already infected), M infects J, destroying it too.罗克斯 3：O 能够感染 Q（以及已被感染的 N），M 感染 J 并将其摧毁。
Day 6: RoXX3 N can only infect destroyed (H and M) or already infected computers (L)第 6 天：RoXX3 N 只能感染已被摧毁（H 和 M）或已受感染的计算机（L）
As neither StuxNet or RoXX3 can make new infections, the computers infected by RoXX3 are: BAEHLNFPOMJQ (roughly in order of infection) so 12 got the virus. Of these ABFHMJ are destroyed (6) leaving 6, infected with RoXX3 but operational. (E,L.P,N,O,Q)由于“震网”病毒和“RoXX3”都无法造成新的感染，被“RoXX3”感染的计算机依次为：BAEHLNFPOMJQ（大致按照感染顺序排列），所以共有 12 台计算机感染了病毒。其中 ABFHMJ 这 6 台已被破坏，剩下 6 台（E、L、P、N、O、Q）虽感染了“RoXX3”但仍在运行。
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You are asked to arrange four tiles into a 2 x 2 square, according to the following rule:要求您根据以下规则将四块瓷砖拼成一个 2×2 的正方形：
Example: 例子:
You are now given the five tiles below.现在给你下面这五块拼图。
You must arrange four of these five tiles into a 2 x 2 square that follows the rule above. (In this case, there is only one possible choice of four tiles that allows this.)您必须从这五块拼图中挑选四块，拼成一个符合上述规则的 2×2 正方形。（在这种情况下，只有唯一一组四块拼图能够做到这一点。）
Question:问题:
Which tile will you NOT use? 你不会用哪块瓷砖？
A. 
B. 
C. 
D.
E. 
Explanation 解释
The correct answer is: 正确答案是：
We assign letters to the tiles as follows:我们按如下方式给这些方块分配字母：
It is possible to create a 2 x 2 square that follows the rule using tiles A, B, D and E, as shown below:可以使用 A、B、D 和 E 这些瓷砖拼出一个符合规则的 2×2 正方形，如下图所示：
Trying the many possibilities until one happens to fit is an option, but there is a way to reduce the number of possibilities that have to be checked.尝试各种可能性直到恰好有一个合适是一种选择，但有一种方法可以减少需要检查的可能性的数量。
We do this in two separate steps.我们分两步来做这件事。
Diagram Explanation: C and E are connected by a line, because both tiles have a little black cloud on one of their sides. A and C are not connected, because none of the symbols on A occurs on C. (The simplified diagram on the right is called a graph.)图示说明：C 和 E 由一条线相连，因为这两块瓷砖中各有一面带有小黑云图案。A 和 C 不相连，因为 A 上的图案在 C 上一个都没有。 （右边的简化图称为图。）
Explanation of the term 4-cycle: a path from tile to tile that follows the lines and ends up at the start after 4 steps from one tile to the next. It is important to realize that the four tiles in a 2 x 2 square that follows the rule, correspond to a 4-cycle.术语“4 循环”的解释：沿着线条从一块瓷砖移动到另一块瓷砖，经过 4 步后回到起点的路径。重要的是要认识到，遵循规则的 2×2 方形中的四块瓷砖对应于一个 4 循环。
Of course, all cyclic permutations of the five letters (i.e. starting from any letter, but keeping the order) are equivalent, as well as their reverse. For example, A-B-C-E-A, B-C-E-A-B and E-C-B-A-E are equivalent.当然，这五个字母的所有循环排列（即从任何字母开始，但保持顺序）都是等价的，它们的逆序排列也等价。例如，A-B-C-E-A、B-C-E-A-B 和 E-C-B-A-E 都是等价的。
Is it possible that the second or third 4-cycle may lead to another solution of our puzzle?第二个或第三个四循环有可能会给我们这个谜题带来另一种解法吗？
Both cycles contain B-C-E, and there are only two possibilities to join these tiles in that order:这两个循环都包含 B-C-E，而且按照这个顺序将这些牌连接起来只有两种可能性：
Neither tile A nor D can be used to complete these to a 2 x 2 square (and still conform to the rule).无论是瓷砖 A 还是 D 都无法用来完成一个 2×2 的正方形（并且仍符合规则）。
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The Beaver Games are an annual competition consisting of three rounds. The beavers always play individually, but in the team rounds (round 1 and 2), the individuals' scores are added together to get a team score.海狸运动会是一项年度赛事，由三轮比赛组成。海狸们总是单独参赛，但在团队赛（第一轮和第二轮）中，个人得分会累加起来得出团队总分。
The teams in round 1 consist of four beavers and are selected by lottery. The team with the highest sum of points wins and advances to the next round. In round 2, new teams of two are selected, again by lottery, from the four beavers who advanced from round 1. The final round is a head-to-head between the two beavers from the winning team from round 2. If you win three rounds in a row, you are the overall winner.第一轮的队伍由四只海狸组成，通过抽签决定。总分最高的队伍获胜并晋级下一轮。第二轮中，从第一轮晋级的四只海狸中再次通过抽签选出两两一组的新队伍。决赛是第二轮获胜队伍中的两只海狸之间的对决。如果你连续赢得三轮，你就是最终的获胜者。
To make sure everyone has fun, even the beavers who do not advance to the next round still do all the activities and their scores are recorded so that next year they can aim for a PB (personal best score).为了确保每个人都能玩得开心，即使那些未能晋级下一轮的海狸也参与所有活动，他们的得分会被记录下来，这样明年他们就可以争取个人最好成绩（PB）。
Beaver Ada 
was the lucky winner of this year's Beaver Games.海狸阿达是今年海狸运动会的幸运冠军。
For each challenge, the individual points are given in the table below.对于每个挑战，其对应的个人得分见下表。
| Beaver: 海狸: | |
 |
 |
 |
 |
 |
 |
 |
| Round 1 第一轮 | 15 | 16 | 19 | 18 | 17 | 20 | 19 | 19 |
| Round 2 第二轮 | 20 | 27 | 30 | 24 | 28 | 24 | 30 | 30 |
| Round 3 第三轮 | 10 | 14 | 11 | 15 | 16 | 13 | 9 | 12 |
Question 问题:
Which three beavers were in Ada's first team?阿达的第一组里有哪三只海狸？
(Drag and drop them onto the question marks. Press 'Save' when you have finished.)（将它们拖放到问号处。完成后按“保存”。）
Explanation 解释
Answer: 答案:
Ada's team in round 1:艾达在第一轮的团队成员：
Explanation: 解释:
The third challenge is an individual match. Since Beaver G is the only player who has fewer points than Ada, they must have been on the same team in the second challenge.第三个挑战是单人赛。由于海狸 G 是唯一一个得分比艾达少的选手，所以他们在第二个挑战中肯定是在同一队。
Their sum of points in the second challenge is 50. This must be higher than the sum of points of the other two-person-team. The two players D and F are the only other ones whose sum of points is less than 50. Therefore, they have to have been in the same team as Ada and G in the first challenge. Since we now know these members, we actually don't have to check the members of the other team.他们在第二项挑战中的总分是 50 分。这必须高于另一支两人队伍的总分。D 和 F 是仅有的总分低于 50 分的另外两名选手。因此，他们肯定在第一项挑战中与艾达和 G 在同一队。既然我们现在已经知道了这些队员，实际上就不必再去查看另一支队伍的队员了。
So this means that in the first challenge the team of (Ada, D, F and G) has 72 points while (B, C, E and H) had 71 points. So Ada's team wins. In the second challenge, (Ada and G) had 50 points and (D and F) had 48 points. In the third challenge, Ada wins with 10 points against G with 9 points. So Ada is the overall winner.这意味着在第一个挑战中，（阿达、D、F 和 G）队获得了 72 分，而（B、C、E 和 H）队获得了 71 分。所以阿达的队伍获胜。在第二个挑战中，（阿达和 G）获得了 50 分，（D 和 F）获得了 48 分。在第三个挑战中，阿达以 10 分战胜了 G 的 9 分。所以阿达是总胜者。
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（Bebras每日一练部分展示）
Ann and Bob are playing on the beach.安和鲍勃正在海滩上玩耍。
Ann has collected many light colored shells, Bob has collected many dark colored pebbles.安收集了许多浅色的贝壳，鲍勃收集了许多深色的卵石。
They made holes in the sand and connected them with furrows.他们在沙地上挖了一些洞，并用沟渠把它们连了起来。
They then take turns to put one of their items in an empty hole.然后他们轮流将自己的一件物品放入一个空洞中。
The loser is the first player to place two of their pieces into two holes connected by a furrow.输家是第一个将两枚棋子放入由一条犁沟相连的两个洞中的玩家。
Ann started the game. They have reached the position shown in the figure below. It is Ann's turn.安开始了游戏。他们已到达如下图所示的位置。现在轮到安了。
Question:问题:
In which empty hole must Ann place a shell to ensure victory for herself?安必须把贝壳放在哪个空洞里才能确保自己获胜？
(Click on it and then press 'Save.')（点击它，然后按“保存”。）
Explanation 解释
Answer: 答案:
Hole number 7. 7号洞。
Explanation: 解释:
Holes 1, 3, 4 and 6 are "closed" to Ann; holes 1, 4, 5 and 6 to Bob. After Anna has put a shell in 7, Bob can play either 2 or 3; in both cases, Ann just puts a shell on 5 to force Bob to lose.对于安来说，1、3、4 和 6 号洞是“封闭”的；对于鲍勃来说，1、4、5 和 6 号洞是“封闭”的。安娜把贝壳放在 7 号洞后，鲍勃可以放 2 号洞或者 3 号洞；无论哪种情况，安只需把贝壳放在 5 号洞就能迫使鲍勃输。
If, from the position shown in the figure, Ann had put a shell in 2, the game would have continued in this way: Bob 7, Ann 5, Bob 3, and Ann loses.如果从图中所示的局面开始，安在 2 位置放了一枚贝壳，那么游戏会这样继续：鲍勃放 7，安放 5，鲍勃放 3，安就输了。
Finally, if Ann had put a shell in 5, the game would have continued in this way: Bob 7, Ann 2, Bob 3 and Ann loses again.最后，如果安在 5 这个位置放了一枚贝壳，游戏就会这样继续：鲍勃 7，安 2，鲍勃 3，安再次输掉。
It is interesting to note that, in the position shown in the figure, if it were Bob's turn, Bob would lose anyway. In fact, if Bob occupies either 2 or 3, Ann occupies 7; instead, if Bob occupies 7, Ann occupies 2, then Bob 3 and Ann 5.有趣的是，在图中所示的位置，如果轮到鲍勃走，鲍勃无论如何都会输。实际上，如果鲍勃占据 2 或 3，安就会占据 7；相反，如果鲍勃占据 7，安占据 2，那么鲍勃占据 3，安占据 5。
But there is another fact, even more interesting: from the initial (empty) board, Bob - who plays second - had a winning strategy... but he made at least one mistake! If Ann's first move was the shell on the left, Bob's top pebble was good (while the bottom one led to a draw) and the mistake occurred on his next move. If, on the other hand, Ann started with the shell on the right, both of Bob's moves were wrong... The task of completing the analysis is yours!但还有一个更有趣的事实：从初始（空的）棋盘开始，后手的鲍勃其实有必胜策略……但他至少犯了一个错误！如果安的第一个棋步是左边的贝壳，鲍勃把石子放在上面是好的（而放在下面则会平局），错误出现在他的下一步。相反，如果安从右边的贝壳开始，鲍勃的两步棋都是错的……完成分析的任务就交给你了！
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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：
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
Answer: 答案:
Explanation: 解释:
There are various ways of solving this task. The most laborious would be to to step through every cell and decide waht colour they should be. Then compare your patterns with those given.解决此任务的方法多种多样。最费力的一种是逐个检查每个单元格，并决定它们应该是什么颜色。然后将你的图案与给出的图案进行比较。
A speedier pattern would be to identify the easiest program to follow,. That is C. It is quick to identify that this must produce the bottom pattern.更快捷的方法是找出最容易遵循的程序，那就是 C。很快就能确定这必然产生最下面的图案。
The three patterns left all contain a different coloured top left cell so now we can apply the rules for each program for just this cell to find out what colour they should be.剩下的三种模式左上角的单元格颜色各不相同，所以现在我们可以针对每个程序仅就这个单元格应用规则，以确定它们应该是什么颜色。
When you are thinking like this, and finding the minimum amount of information required to solve a problem while ignoring other information, you are using the computational thinking skill Abstraction.当你这样思考时，在解决问题时寻找所需的最少信息而忽略其他信息，你就是在运用计算思维技能中的抽象能力。
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（Bebras每日一练部分展示）
2022-2023年Bebras挑战练习题
Juniors (14-16岁) 集合
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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。 |
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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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