Codeforces Round #337 (Div. 2) C. Harmony Analysis 构造

C. Harmony Analysis

题目连接：

http://www.codeforces.com/contest/610/problem/C

Description

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 Input

2

Sample Output

++**

++

++++

+**+

Hint

题意

要求你构造出2^n个2n维向量，使得向量之间两两相乘都等于0

题解：

瞎构造的。。。

大概证明可以由数学归纳法证明

假设我现在已经构造出了

a

那么我就可以构造出

a a

a -a

然后一直重复就好了。。。

代码

#include<bits/stdc++.h>
using namespace std;

int dp[1200][1200];
int n;
int main()
{
    scanf("%d",&n);
    dp[0][0]=1;
    for(int x=1;x<=n;x++)
    {
        for(int i=0;i<(1<<x-1);i++)
        {
            for(int j=0;j<(1<<x-1);j++)
            {
                dp[i][j+(1<<x-1)]=dp[i][j];
                dp[i+(1<<x-1)][j]=dp[i][j];
                dp[i+(1<<x-1)][j+(1<<x-1)]=1-dp[i][j];
            }
        }
    }
    for(int i=0;i<(1<<n);i++)
    {
        for(int j=0;j<(1<<n);j++)
        {
            if(dp[i][j])printf("+");
            else printf("*");
        }
        printf("\n");
    }
    return 0;
}