Codeforces - 337C(div2) - 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.

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

嘿嘿嘿，找规律题，so happy

本来2^9维光想想就要命，根本不知道怎么玩= =虽然多种情况任意输出，但是感觉一种就够我炸了= =

然后自己画了画........0 0？.........0 0！2^1维和2^2维画了几种2333尼玛就是个找规律啊，摔！

规律：答案分成四部分，左上部分 = 右上部分= 左下部分，右下部分跟他们相反（应该也有其他规律，不过懒得看了= =）（我才不说我是靠2^1维2^2维强行猜的规律= =）

求2^(k-1)维扩展到2^k维，所以从2^1维扩展到2^9维，进行预处理

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <stdlib.h>
#include <cmath>
#include <cstring>
#include <string.h>
#include <cstdlib>
#include <iostream>
#include <set>
using namespace std;

int a[550][550];

int main()
{
int n;
memset(a, 0, sizeof(a));
a[0][0] = 1;
int cnt = 1;
int sum = 9;
while(sum--){
for(int i = 0; i < cnt; i++){
for(int j = 0; j < cnt; j++){
a[i + cnt][j] = a[i][j + cnt] = a[i][j];
a[i + cnt][j + cnt] = -a[i][j];
}
}
cnt *= 2;
}
while(scanf("%d", &n) != EOF){
int ans = 1;
for(int i = 0; i < n; i++)
ans *= 2;
for(int i = 0; i < ans; i++){
for(int j = 0; j < ans; j++){
if(a[i][j] == 1)
printf("+");
else
printf("*");
}
printf("\n");
}
}
return 0;
}