Javascript 以JS格式将纸张切割成最少的正方形

我想把算法从 在javascript中

我能够轻松地在js（）中翻译第一个算法，但是贪婪的方法不够精确

我在js和递归编程方面遇到了一些问题，在我第一次尝试时，我得到了最大调用堆栈大小超出错误，因此我尝试使用

节点--stack size=10000

，但在这种情况下，脚本没有输出任何内容

这就是我现在拥有的：

// JS program to find minimum  
// number of squares  
// to cut a paper using Dynamic Programming  

const MAX = 300


let dp = new Array(MAX);

for (let i = 0; i < dp.length; i++) {
  dp[i] = new Array(MAX).fill(0);
}

// Returns min number of squares needed  
function minimumSquare(m, n) {

    // Initializing max values to  
    // vertical_min  
    // and horizontal_min 
    let vertical_min = 10000000000
    let horizontal_min = 10000000000


    // If the given rectangle is  
    // already a square  
    if (m === n) {
        return 1
    }

    // If the answer for the given rectangle is  
    // previously calculated return that answer  
    if (dp[m][n] !== 0) {
        return dp[m][n]
    }

    // The rectangle is cut horizontally and  
    // vertically into two parts and the cut  
    // with minimum value is found for every  
    // recursive call.  
    for (let i=1; i<m/2+1; i++) {
        // Calculating the minimum answer for the  
        // rectangles with width equal to n and length  
        // less than m for finding the cut point for  
        // the minimum answer 
        horizontal_min = Math.min(minimumSquare(i, n) + minimumSquare(m-i, n), horizontal_min)

    }

    for (let j=1; j<n/2+1; j++) {

        // Calculating the minimum answer for the  
        // rectangles with width equal to n and length  
        // less than m for finding the cut point for  
        // the minimum answer 
        vertical_min = Math.min(minimumSquare(m, j) + minimumSquare(m, n-j), vertical_min)

    }

    // Minimum of the vertical cut or horizontal 
    // cut to form a square is the answer 
    dp[m][n] = Math.min(vertical_min, horizontal_min) 
    return dp[m][n]

}

// Driver code 
    let m = 30
    let n = 35
    console.log(minimumSquare(m, n))

//This code is contributed by sahilshelangia 

//JS程序找到最小值
//方块数
//用动态规划法剪纸
常数最大值=300
设dp=新阵列（最大值）；
for（设i=0；i对于（设i=1；i我添加了两个新条件来解决问题


如果m可被n整除，则返回值m/n
如果n可被m整除，则返回值n/m


const MAX=300
设dp=新阵列（最大值）；
for（设i=0；i为了（让我=1；我为没有尝试这个而感到愚蠢！非常感谢！