在 JavaScript 中计算两个或三个数字的 LCM

Question

我正在使用以下代码来确定两个或三个数字的最大公分母 (GCD)：

Math.GCD = function(numbers) {
  for (var i = 1 ; i < numbers.length ; i++){
    if (numbers[i] || numbers[i] === 0)
      numbers[0] = twogcd(numbers[0], numbers[i]);
  }
  return numbers[0];

  function twogcd(first, second) {
    if (first < 0) first = -first;
    if (second < 0) second = -second;
    if (second > first) {var temp = first; first = second; second = temp;}
    while (true) {
        first %= second;
        if (first == 0) return second;
        second %= first;
        if (second == 0) return first;
    }
   }
};

Math.LCM = function(first,second) {
    return first * (second / this.GCD(first, second)); // CANNOT FIGURE OUT HOW TO EXTEND THIS TO THREE #s
};

// example
console.log(Math.GCD([4, 5, 10]));

Demo also in JSFiddle

注意其中关于最小公倍数 (LCM)

的函数

我正在尝试扩展此函数，以便它可以计算相同的两个或三个用户提供的输入的 LCM，但我终其一生都无法做到正确。我是 JavaScript 的新手，如果有任何帮助，我将不胜感激。请注意，如果某个字段留空，则也应将其从计算中省略，就像 GCD 所做的那样。

Answer 1

您可以使用这些功能：

function gcd2(a, b) {
  // Greatest common divisor of 2 integers
  if(!b) return b===0 ? a : NaN;
  return gcd2(b, a%b);
}
function gcd(array) {
  // Greatest common divisor of a list of integers
  var n = 0;
  for(var i=0; i<array.length; ++i)
    n = gcd2(array[i], n);
  return n;
}
function lcm2(a, b) {
  // Least common multiple of 2 integers
  return a*b / gcd2(a, b);
}
function lcm(array) {
  // Least common multiple of a list of integers
  var n = 1;
  for(var i=0; i<array.length; ++i)
    n = lcm2(array[i], n);
  return n;
}

Answer 2

我还没弄清楚你的代码在做什么，但这里是查找 LCM 的函数：

LCM = function(numbers) {
  console.log(numbers)  
  if (numbers.length < 2) return
  first = numbers[0]
  second = numbers[1]
  var i = j = 1
  var mult1 = first * i++
  var mult2 = second * j++
  while (mult1 != mult2) {
    if (mult1 < mult2)
        mult1 = first * i++
    else
        mult2 = second * j++
  }
  if (numbers.length > 2) {
    numbers[1] = mult1 //I hope you're fine with the fact that 'numbers' gets modified
    mult1 = LCM(numbers.splice(1, numbers.length-1))
  }
    return mult1
}

我知道它效率不高，但它说明了如何将它与任意数量的参数一起使用（它只是递归调用）。

小提琴：https://jsfiddle.net/grabantot/fr0gzogL/

Answer 3

也许你稍微改变GCD和LCM的结构，让这两个方法只有两个参数。

要获得两个以上参数的结果，请使用Array.prototype.reduce()，它从数组中获取两个元素并返回一个结果，该结果用作新的插入，直到数组完成。

虽然LCM 和GCD 是associative，但您可以根据需要将其链接起来。

Math.GCD = function twogcd(first, second) {
    if (first < 0) first = -first;
    if (second < 0) second = -second;
    if (second > first) { var temp = first; first = second; second = temp; }
    while (true) {
        first %= second;
        if (first == 0) return second;
        second %= first;
        if (second == 0) return first;
    }
};

Math.LCM = function (first, second) {
    return first * (second / Math.GCD(first, second));
};

document.getElementById('calc').addEventListener('click', function (e) {
    var first = +document.getElementById("first").value,
        second = +document.getElementById("second").value,
        third = +document.getElementById("third").value,
        numbers = [first, second, third],
        resultGCD = numbers.reduce(Math.GCD), // just chain it together
        resultLCM = numbers.reduce(Math.LCM); // just chain it together

    document.getElementById('gcd').innerHTML = resultGCD;
    document.getElementById('lcm').innerHTML = resultLCM;
});

GCD: <span id="gcd"></span><br />
LCM: <span id="lcm"></span><br />
<form name="sci-calc" method="POST" id="sci-calc">
    <input type="text" name="stuff[]" class="input-field" id="first" /><br />
    <input type="text" name="stuff[]" class="input-field" id="second" /><br />
    <input type="text" name="stuff[]" class="input-field" id="third" /><br />
    <button type="button" id="calc">CALC</button>
</form>