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2^n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("
");
}
return 0;
}