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;
}