3
7
2016
0

[Codeforces Round #337] Harmony Analysis


C. Harmony Analysis
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

The semester is already ending, so Danil made an effort and decided to visit a lesson on harmony analysis to know how does the professor look like, at least. Danil was very bored on this lesson until the teacher gave the group a simple task: find 4 vectors in 4-dimensional space, such that every coordinate of every vector is 1 or - 1 and any two vectors are orthogonal. Just as a reminder, two vectors in n-dimensional space are considered to be orthogonal if and only if their scalar product is equal to zero, that is:

.

Danil quickly managed to come up with the solution for this problem and the teacher noticed that the problem can be solved in a more general case for 2k vectors in 2k-dimensinoal space. When Danil came home, he quickly came up with the solution for this problem. Can you cope with it?

Input

The only line of the input contains a single integer k (0 ≤ k ≤ 9).

Output

Print 2k lines consisting of 2k characters each. The j-th character of the i-th line must be equal to ' * ' if the j-th coordinate of the i-th vector is equal to  - 1, and must be equal to ' + ' if it's equal to  + 1. It's guaranteed that the answer always exists.

If there are many correct answers, print any.

Examples
input
2
output
++**
+*+*
++++
+**+
Note

Consider all scalar products in example:

  • Vectors 1 and 2( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( + 1) + ( - 1)·( - 1) = 0
  • Vectors 1 and 3( + 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) + ( - 1)·( + 1) = 0
  • Vectors 1 and 4( + 1)·( + 1) + ( + 1)·( - 1) + ( - 1)·( - 1) + ( - 1)·( + 1) = 0
  • Vectors 2 and 3( + 1)·( + 1) + ( - 1)·( + 1) + ( + 1)·( + 1) + ( - 1)·( + 1) = 0
  • Vectors 2 and 4( + 1)·( + 1) + ( - 1)·( - 1) + ( + 1)·( - 1) + ( - 1)·( + 1) = 0
  • Vectors 3 and 4( + 1)·( + 1) + ( + 1)·( - 1) + ( + 1)·( - 1) + ( + 1)·( + 1) = 0

题目大意:给出正整数$k$,求一个$2^k\times 2^k$的矩阵$A$,使得$a_{i,j}\in\left\{1,-1\right\},\forall i,j\in\left[1,2^k\right]\left(i\neq j\right)\ \sum_\limits{x=1}^\limits{2^k}a_{i,x}\cdot a_{j,x}=0$。


在打比赛时完全没有思路,于是这场我只做了前两题。。。

根据官方题解:递归求解。对于$k=0$,答案就是$1$或者$-1$;对于$k=i$,答案是把$k=i-1$的答案放在四个角上,并对放在右下方的那个矩阵的所有元素取相反数。

证明什么的我就不管了,因为我觉得这解法完全是shi出来的,没有什么意义。

代码在此。

Category: 题解 | Tags: Codeforces | Read Count: 515

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