这个题目来自我大一上学期编程Online Challenges，我认为是我那个学期经历的最棘手的问题（没有之一）QwQ
本篇文章仅包含题目的内容，感兴趣的朋友一起讨论交流，我会在之后的文章里分享我的思路和代码～
首先是原汁原味的英文:
Bored of sudoku? Try calcudoku!
The puzzle is simple. Upon a square n×n grid, each row and column must contain a permutation of the numbers from 1 to n. Consequently, each row or column must contain each of these ==exactly once==.
Overlaid upon the grid are a number of cages. Each cage consists of one or more contiguous cells, and contains a label. A label is a target value V, and optionally an operator, which may be any one of +, −, ∗, /.
A player must place numbers in every cage such that its indicated condition is satisfied. The rule for a given cage depends on its operator:
- no operator : the cage must consist of a single square containing V;
- ∗ : the product of the numbers in the cage must be equal to V;
- + : the sum of the numbers in the cage must be equal to V;
- / : the cage must be exactly two squares, the smaller value of the two must divide the larger one evenly so that the quotient is V;
- − : the cage must contain two cells, where the difference of the two members is V.
Your task Puzzles will be provided in the following way. Here is an example using the picture above.
0 0 1 2 2 3
0 4 4 5 5 3
6 7 8 8 9 9
6 6 10 10 11 11
12 13 13 14 14 15
16 16 13 14 17 15
80* 3 5- 2/ 11+ 1- 9* 2 3- 30* 11+ 2/ 6 8* 13+ 8+ 10* 1
The first six lines describe the cage shapes. All cells assigned the same number share a common cage.
The seventh line contains the cage labels. Which cage a label belongs to depends on its index in the line. So, 80∗ is the condition for cage 0, while 6 is the label upon cage 12. A label will always be a number, followed optionally by an operator symbol.
You should print your solution to standard output, as follows:
5 4 3 1 6 2
4 6 5 2 3 1
3 2 1 4 5 6
1 3 6 5 2 4
6 1 2 3 4 5
2 5 4 6 1 3
Each row must be on a line by itself. Each number is to be separated from the next by a single space.
If a puzzle is unsolvable, you should output ==No solution.==
Notes
-
All test puzzles will be 6×6.
-
All test puzzles will have a unique solution, if one exists.
-
The first six input lines will each consist of exactly six - -decimal integers separated by spaces.
-
The cage numbers will be numbered from 0 to m, increasing as new cages are encountered, scanning left to right, top to bottom.
-
There will be exactly m+1 labels provided on the 7th line, space-separated as above.
Examples An unsolvable example puzzle:
Input:
0 1 2 2 3 3
4 4 4 5 5 5
4 4 6 6 7 8
9 10 10 11 7 8
9 12 12 13 13 8
14 14 12 13 15 15
6 6 2* 3/ 10+ 6* 6+ 5 24* 2- 2- 2 7+ 20* 8+ 2-
Output:
No solution.
Hints
-
Consider the use of recursion.
-
Functions are your friends. Break the problem up into more manageable chunks.
-
For certain puzzles, direct deduction will only get you so far. At some point you may be forced to make a guess and see if it leads to a violation of the rules.
本人中文翻译：
规则非常简单，在n x n尺寸的表格中各行各列都填入数字1 ～ n，同时保证每行每列都没有重复的数字。
整个表格被粗线分成若干个框框，而每个框框都包含至少一个格子以及一个标签。一个标签又由一个目标值和一个运算符（==当框框里只含有一个格子时，不需要运算符==）组成。
玩家在把数字填入格子中时要使框框的要求满足。而具体的要求，由标签中的运算符决定：
- 没有运算符：这个框框一定只有一个格子，同时格子里填的数一定是标签上的目标值。
- *：框框中各个格子的数字的乘积等于目标值。
- +：框框中各个格子的数字的和等于目标值。
- /：框框中有且只包含两个格子，同时较大的数字能整除较小的数字所得的商等于目标值。
- -：框框中有且只包含两个格子，同时两数的差等于目标值。
你需要做的工作 具体的数独表格会按以下格式输入。以下代码是以图片中的题目为例输入的:
0 0 1 2 2 3
0 4 4 5 5 3
6 7 8 8 9 9
6 6 10 10 11 11
12 13 13 14 14 15
16 16 13 14 17 15
80* 3 5- 2/ 11+ 1- 9* 2 3- 30* 11+ 2/ 6 8* 13+ 8+ 10* 1
前六行的数字，描述了题目中各个框框的具体形状。数字相同的格子代表它们在同一个框框中。 第七行按顺序则相继代表各个框框所对应的标签。比如 80*是0号框框所对应的标签，6 则对于数字12所在的框框。标签的输入格式永远是目标值在前，运算符在后（需要运算符的情况下）。
你需要将求出的答案，按以下格式print出来。
5 4 3 1 6 2
4 6 5 2 3 1
3 2 1 4 5 6
1 3 6 5 2 4
6 1 2 3 4 5
2 5 4 6 1 3
行对行列对列，数字之间以空格隔开。 如果所给题目无解，需要print No solution.
注意事项（本栏下的内容主要是说明我们挑战审核系统里面testcases的一些细节，可忽略）
- 所有testcases的尺寸都是6 x 6。
- 所有testcases在有解的前提下，都有唯一解。
- 所有testcases 前6行的input都是范围内的整数，并以空格隔开。
- 前六行的input框框标号都是从0开始到m，顺序从左到右从上到下。
- 第七行共有m + 1个标签输入，相互以空格隔开。
例子 一个无解的例题 输入：
0 1 2 2 3 3
4 4 4 5 5 5
4 4 6 6 7 8
9 10 10 11 7 8
9 12 12 13 13 8
14 14 12 13 15 15
6 6 2* 3/ 10+ 6* 6+ 5 24* 2- 2- 2 7+ 20* 8+ 2-
输出：
No solution.
提示（学校好良心QwQ）
- 考虑用用递归。
- 自己def函数是一个好办法，尽量把一个复杂的问题拆分成数个部分解决。
- 对于一些情况，直接推导并不能走的太远。有时你可能需要通过猜来推进，再观察结果是否违背了要求。