这是什么样的矩阵？

Question

这是一个奇怪的矩阵。如果输入为1，则输出为1.但如果输入为2，则输出为矩阵3 ^ 2。最奇怪的是输入3然后输出是矩阵5 ^ 2。有谁知道如何解决这个问题？我真的需要找到工作的答案！

Answer 1

这非常简单。矩阵的维数if（2 * input）+1和矩阵中的数字是距离中心+1的距离。请参阅下面的python代码。

import math

def distance(x1, y1, x2, y2):
    dist = (x1-x2)*(x1-x2) + (y1-y2)*(y1-y2)
    return int(math.sqrt(dist))+1

if __name__=='__main__':
    x = int(input())
    if(x < 1):
        exit()
    dimension = (2*x)-1
    center = dimension // 2
    mat = [[None for i in range(dimension)] for i in range(dimension)]
    for i in range(dimension):
        for j in range(dimension):
            mat[i][j] = distance(center, center, i, j);
            print(mat[i][j], end=' ')
        print('')

Answer 2

这是基于您的代码的PHP代码：

<?php   
        $x = 3;
        $dimension = (2 * $x) -1 ;  
        $center = floor($dimension/2); 
        for($i = 1; $i <= $dimension; $i++)
        {
            for($j = 1; $j <= $dimension; $j++)
            { 
                $dist = ($center - $i) * ($center - $i) + ($center - $j) * ($center - $j);
                $result = floor(sqrt($dist)) + 1; 
                echo "(" . $i . ", " . $j . ")" . " = " . $result . "</br>";                
            }       
        }
    ?>

它没有给我与输出相同的结果。

Answer 3

我已经有了如何在PHP中做到这一点，但我不知道如何优化代码。有谁知道如何正确地做到这一点？

<?php   
$z = 3; 
$dimension = (2 * $z) -1 ;  

for($i = 1; $i <= $dimension; $i++)
{
    for($j = 1; $j <= $dimension; $j++)
    { 
        if(($i == 1|| $j == 1) || ($i == $dimension || $j == $dimension)){
            echo $z . " "; 
        } else if(($i == 2 || $j == 2) || ($i == $dimension - 1 || $j == $dimension - 1)){
            echo $z - 1 . " "; 
        }
        else if(($i == 3 || $j == 3) || ($i == $dimension - 2 || $j == $dimension - 2)){
            echo $z - 2 . " "; 
        }
        else if(($i == 4 || $j == 4) || ($i == $dimension - 3 || $j == $dimension - 3)){
            echo $z - 3 . " ";
        }
        else{
            echo $z - 4 . " "; 
        }

        if($j == $dimension){
            echo "</br>"; 
        }
    }       
}   

?>