【53.57%】【codeforces 610C】Harmony Analysis

time limit per test3 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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

【题解】



规律题;

k和k-1的答案存在如下关系;



上图中的方框表示k-1时的答案;

把它按照上述方式复制3份，第4份则在原来的基础上取反;(用1表示+，0表示减);

比如样例输入

++**
+*+*
++++
+**+
//->
1100
1010
1111
1001
//->相同的3份取反的一份
1100 1100
1010 1010
1111 1111
1001 1001

1100 0011
1010 0101
1111 0000
1001 0110
/*而这正是k=3时的答案;右下角那个
    上面两个方框是肯定满足的;
    为了让下面两个方框在乘的时候也满足;
    相当于左边取A，右边取它的相反数-A;这样一减就是0;
    如果一个答案符合要求则全部取反还是能符合要求的;
    所以右下角取反不会影响下面两个的答案正确性;
    然后又能让上面两个方框的乘下面两个方框的向量的时候积为0；所以是符合要求的;
*/

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
#define lson L,m,rt<<1
#define rson m+1,R,rt<<1|1
#define LL long long

using namespace std;

const int MAXN = 1000;
const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
const double pi = acos(-1.0);
struct abc
{
    int a[1000][1000];
};

int k;
abc ans[10];

void input_LL(LL &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t) && t!='-') t = getchar();
    LL sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

void input_int(int &r)
{
    r = 0;
    char t = getchar();
    while (!isdigit(t)&&t!='-') t = getchar();
    int sign = 1;
    if (t == '-')sign = -1;
    while (!isdigit(t)) t = getchar();
    while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
    r = r*sign;
}

int main()
{
   // freopen("F:\\rush.txt","r",stdin);
    ans[0].a[1][1] = 0;
    for (k= 1;k <= 9;k++)
    {
        for (int i = 1;i <= 1<<(k-1);i++)
            for (int j = 1;j <= 1<<(k-1);j++)
            {
                ans[k].a[i][j] = ans[k-1].a[i][j];
                ans[k].a[i+(1<<(k-1))][j] = ans[k-1].a[i][j];
                ans[k].a[i][j+(1<<(k-1))] = ans[k].a[i][j];
                ans[k].a[i+(1<<(k-1))][j+(1<<(k-1))] = !ans[k].a[i][j];
            }
    }
    input_int(k);
    for (int i = 1;i <= 1<<k;i++)
    {
        for (int j = 1;j <= 1<<k;j++)
            if (ans[k].a[i][j])
                putchar('+');
            else
                putchar('*');
        puts("");
    }
    return 0;
}