codeforces 610C Harmony Analysis 构造

题目如下：

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

++**
+*+*
++++
+**+

思路：这道题的关键在于要能发现n=i(i>0)的结果，可以由n=i-1的结果构造出来。即如果n=i-1时为A，那么n=i时，可以构造A A/A -A(-A就是A中每个取相反的符号,/表示换一行)。可以证明这是符合要求的。这样一来，这道题就简单了。

代码如下：

#include

using namespace std;

const int maxn = 1<<11;

char s[maxn][maxn];

char op(char c){

    if (c == '*') {

        return '+';

    }else return '*';

}

int main(int argc, const char * argv[]) {

    int n;

    cin >> n;

    s[1][0] = '+';s[1][1] = '\0';

    int l = 1;

    for (int i=0; i

        for (int j=1; j<=l; j++) {

            for (int k=0; k

                s[j][k+l] = s[j][k];

            }

            s[j][l*2] = '\0';

        }

        for (int j=l+1; j<=2*l; j++) {

            for (int k=0; k

                s[j][k] = s[j-l][k];

                s[j][k+l] = op(s[j][k]);

            }

            s[j][l*2] = '\0';

        }

        l*=2;

    }

    for (int i=1; i<=l; i++) {

        cout << s[i] << endl;

    }

    return 0;

}