我有一个数字，我需要分成5个部分。但是，我希望每个部分都是一个随机数。但是当所有部件加在一起时，它们等于原始数字。我不确定如何使用JavaScript执行此操作。此外，我不希望分割部分的min为0或1，我想自己设置min。
例如，数字是450.我希望分割的部分不小于60.所以首先，数组将是[60,60,60,60,60]。但是我想随机化，以便它们总计达到450.这样做最好的方法是什么？
谢谢！
这是我到目前为止所尝试的：
let i = 0;
let number = 450;
let numArray = [];
while(i <= 5){
while(number > 0) {
let randomNum = Math.round(Math.random() * number) + 1;
numArray.push(randomNum);
number -= randomNum;
}
i += 1;
}
答案 0 :(得分：3)
将五个最小值（例如min = 60）加起来：
var minSum = 5 * min
然后获得原始号码（orNumber = 450）和minSum之间的差异。
var delta = orNumber - minSum
现在您可以获得4个不同的随机数，范围从0到独占1.
将这些数字按升序排序。
这些randoms的主要做法如下：
delta并获得其中一个部分。最后一部分是delta减去所有其他部分。
之后您只需将min添加到所有部分。
答案 1 :(得分：2)
让你的数字为N，让p n 为第n部分。获得5个部分：
修改
为了让它看起来更随机，在生成数字后随机播放
答案 2 :(得分：2)
此函数生成从0到1的随机数，将它们相加，以确定需要乘以它们以提供正确的范围。它的好处是所有数字都是公平分配的。
function divvy(number, parts, min) {
var randombit = number - min * parts;
var out = [];
for (var i=0; i < parts; i++) {
out.push(Math.random());
}
var mult = randombit / out.reduce(function (a,b) {return a+b;});
return out.map(function (el) { return el * mult + min; });
}
var d = divvy(450, 6, 60)
console.log(d);
console.log("sum - " + d.reduce(function(a,b){return a+b}));
答案 3 :(得分：1)
您可以使用do..while循环从原始数字中减去最小数字，在循环结束时保留原始数字的副本以将剩余部分推送到数组
let [n, total, m = n] = [450, 0];
const [min, arr] = [60, []];
do {
n -= min; // subtract `min` from `n`
arr.push(n > min ? min : m - total); // push `min` or remainder
total += arr[arr.length - 1]; // keep track of total
} while (n > min);
console.log(arr);
要在结果数组中随机化输出，请选择大于min且小于n的数字，以在特定范围内创建随机数
let [n, total, m = n] = [450, 0];
const [min, arr, range = min + min / 2] = [60, []];
do {
let r = Math.random() * (range - min) + min; // random number in our range
n -= r; // subtract `min` from `n`
arr.push(n > min ? r : m - total); // push `r` or remainder
total += arr[arr.length - 1]; // keep track of total
} while (n > min);
console.log(arr);
答案 4 :(得分：0)
我为初学者制作了更长版本。
const n = 450;
const iterations = 5;
const parts = [];
// we'll use this to store what's left on each iteration
let remainder = n;
for (let i = 1; i <= iterations; i += 1) {
// if it's the last iteration, we should just use whatever
// is left after removing all the other random numbers
// from our 450
if (i === iterations) {
parts.push(remainder);
break;
}
// every time we loop, a random number is created.
// on the first iteration, the remainder is still 450
const part = Math.round(Math.random() * remainder);
parts.push(part);
// we must store how much is left after our random numbers
// are deducted from our 450. we will use the lower number
// to calculate the next random number
remainder -= part;
}
// let's print out the array and the proof it still adds up
const total = totalFromParts(parts);
console.log(parts);
console.log('Total is still ' + total);
// this function loops through each array item, and adds it to the last
// just here to test the result
function totalFromParts(parts) {
return parts.reduce((sum, value) => sum + value, 0);
}
答案 5 :(得分：0)
有更有效的方法对此进行编码，但为了解释解决问题的逻辑，这将逐步介绍，转换价值并解释逻辑。
// Set start number, number of fragments
// minimum fragment size, define fragments array
var n = 450
var x = 5
var minNumber = 60
var fragment = n / x
// stuff array with equal sized fragment values
var fragments = []
for (i = 0; i < x; i++) {
fragments[i] = fragment;
}
document.write("fragments: " + fragments);
var delta = [];
// iterate through fragments array
// get a random number each time between the fragment size
// and the minimum fragment sized defined above
// for even array slots, subtract the value from the fragment
// for odd array slots, add the value to the fragment
// skip the first [0] value
for (i = 1; i< x; i++) {
delta[i] = Math.floor(Math.random() * (fragment - minNumber));
document.write("<br />delta: " + delta[i]);
if((i % 2) == 1) {
fragments[i] -= delta[i]
}
else {
fragments[i] += delta[i]
}
}
// set the initial fragment value to 0
fragments[0] = 0
// defines a function we can use to total the array values
function getSum(total, num) {
return total + num;
}
// get the total of the array values, remembering the first is 0
var partialTotal = fragments.reduce(getSum)
document.write("<br />partial sum: " + partialTotal);
// set the first array value to the difference between
// the total of all the other array values and the original
// number the array was to sum up to
fragments[0] = (n - partialTotal)
// write the values out and profit.
document.write("<br />fragments: " + fragments);
var grandTotal = fragments.reduce(getSum)
document.write("<br />Grand total: " + grandTotal);