Codeforces Round #337 (Div. 2) C 找规律+构造

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C. 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

题意：

+表示1，*表示-1，找出满足条件的n维坐标系下所有向量的乘积为0

思路：

写两组样例就看出来了

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<string>#include<vector>#include <ctime>#include<queue>#include<set>#include<map>#include<list>#include<stack>#include<iomanip>#include<cmath>#include<bitset>#define mst(ss,b) memset((ss),(b),sizeof(ss))///#pragma comment(linker, "/STACK:102400000,102400000")typedef long long ll;typedef long double ld;#define INF (1ll<<60)-1#define Max 1e9using namespace std;int n;int p[10];char a[10][555][555];int main(){    p[0]=1;p[1]=2;    a[0][1][1]='*';    a[1][1][1]=a[1][1][2]=a[1][2][1]='+';a[1][2][2]='*';    for(int k=2;k<=9;k++){        p[k]=p[k-1]*2;        for(int i=1;i<=p[k-1];i++){            for(int j=1;j<=p[k-1];j++){                a[k][i][j]=a[k-1][i][j];            }        }        for(int i=1;i<=p[k-1];i++){            for(int j=p[k-1]+1;j<=p[k];j++){                a[k][i][j]=a[k-1][i][j-p[k-1]];            }        }        for(int i=p[k-1]+1;i<=p[k];i++){            for(int j=1;j<=p[k-1];j++){                a[k][i][j]=a[k-1][i-p[k-1]][j];            }        }        for(int i=p[k-1]+1;i<=p[k];i++){            for(int j=p[k-1]+1;j<=p[k];j++){                if(a[k-1][i-p[k-1]][j-p[k-1]]=='+')                    a[k][i][j]='*';                else a[k][i][j]='+';            }        }    }    scanf("%d",&n);    for(int i=1;i<=p[n];i++){        for(int j=1;j<=p[n];j++){            printf("%c",a[n][i][j]);        }        printf("\n");    }    return 0;}