Codeforces 610C：Harmony Analysis 递归

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.

Sample test(s)

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

题意是给出2^k次方个2^k维的互相叉乘都是0的向量。

如果要求2^k次方个，可以先考虑2^(k-1)，将其按照矩形的方式复制四份，然后最后的那一份取反。

证明也很简单，上边没有取反的那部分肯定互相之间是叉乘是0的，因为前半部分叉乘是0，后半部分叉乘也是0。

上下部分，如果是原来的自己取反的，那么前半部分和后半部分的相乘一定是互为相反数的关系。如果不是自己取反的那一个，那就是其它的后半部分取反的，没有影响，还是0。

递归求解。

代码：

#pragma warning(disable:4996)#include <iostream>#include <algorithm>#include <cmath>#include <vector>#include <string>#include <cstring>#include <set>#include <queue>#include <stack>using namespace std;const int maxn = 530;int k;int val[maxn][maxn];void rec(int n, int x, int y, int v){if (n == 1){val[x][y] = v;}else{rec(n / 2, x, y, v);rec(n / 2, x + n / 2, y, v);rec(n / 2, x, y + n / 2, v);rec(n / 2, x + n / 2, y + n / 2, -v);}}void input(){cin >> k;}void solve(){rec(1 << k, 1, 1, 1);int i, j;for (i = 1; i <= (1 << k); i++){for (j = 1; j <= (1 << k); j++){if (val[i][j] == 1){cout << "+";}else{cout << "*";}}cout << endl;}}int main(){//freopen("i.txt","r",stdin);//freopen("o.txt","w",stdout);input();solve();//system("pause");return 0;}