Codeforces 610С — Harmony Analysis 找规律

Codeforces Round #337 (Div. 2)

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: find4 vectors in 4-dimensional space, such that every coordinate of every vector is1 or  - 1 and any two vectors are orthogonal. Just as a reminder, two vectors inn-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 for2^k vectors in2^k-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 2^k lines consisting of2^k characters each. Thej-th character of the i-th line must be equal to ' * ' if thej-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

分析：把上面那个变化一下，就可以看出规律了。

2
++   ++
+*    +*

++    **
+*    *+

可以看出左上，右上和左下三部分是相同的，右下部分是和其他三部分相反的。

#include<bits/stdc++.h>#define LL long longusing namespace std;int a[1025][1025];int main(){    int n;    cin>>n;    int t=1;    a[0][0]=1;    while(n--){        for(int i=0;i<t;i++){            for(int j=0;j<t;j++){                a[i+t][j]=a[i][j+t]=a[i][j];                a[i+t][j+t]=-a[i][j];            }        }        t<<=1;    }    for(int i=0;i<t;i++){        for(int j=0;j<t;j++){            if(a[i][j]==1)putchar('+');            else putchar('*');        }        printf("\n");    }    return 0;}