Question

我已经完成了背包问题，但是我不确定这个问题是否会在某个方面落在那个区域，无论如何，如何解决它。这是一个问题：

Suppose we are given a set of denominations {1, 2, 10, 20, 50, 100}. A customer is checking out and is paying in cash. How we suggest the top 6 or top x choices on our own payment screen based on what amount the customer may have handed over. For example if customer has to pay 87$, customer might hand over 5 bills of 100, or 4 bills of 20 and a 5 and a 2, etc我无法想出一个算法，虽然我一直觉得这是围绕背包问题，但对给定值有多个答案？

Answer 1

值得一提的是，您应该获得更具体的要求。与客户坐在一起，准备示例并向他们提问是非常重要的：＆＃34;我必须支付87，我给你87x $ 1，是不是好，为什么＆＃34;。从更明显的角度来看，你得到：＆＃34;什么是更好的50x2或50 + 20 + 20以及为什么＆＃34;。要创建任何算法，首先需要说明你将如何以人为本（无限时间和耐心）然后你可以实现它。

这里一个非常简单的指标是，您希望价值尽可能接近实际金额，而且您不想用太多账单支付，对吗？

所以我们必须找到所有的解决方案，然后计算得分（更少的账单和更接近目标值的更好），按分数排序并采取＆＃39; x＆＃39;最好的。

你可以升级＆＃34;得分＆＃34;功能与＆＃34; 10和20＆＃34;账单是首选，所以他们使用它们给你更好的分数。或者，较低的账单金额比目标价值与账单金额之间的差异更重要，因此您会更多地惩罚账单（比如每10美元的差额与1个账单的罚款相同）

这是基于回溯的javascript代码。

＆＃13;

// const countables = [100, 50, 20, 10, 2, 1];
// const targetAmount = 87;

const countables = [5, 2, 1];
const targetAmount = 13;


const payments = [];
const solutions = [];
let index = 0;

let end = false;
let i = 0;

while (end === false) {
    const sum = payments.reduce((prev, val) => prev + val, 0);
    if (sum >= targetAmount) {
        solutions.push({
            sum,
            score: countTheScore(payments, targetAmount, sum),
            payments: [...payments],
        })
        index++;
        if (index === countables.length) {
            let popped = null;
            do {
                popped = payments.pop();
            } while (popped === countables[countables.length - 1]);
            if (payments.length === 0) {
                end = true;
            }
            const indexOfLastNonLowest = countables.indexOf(popped);
            index = indexOfLastNonLowest + 1;
        } else {
            payments.pop();
        }
    } else {
        payments.push(countables[index]);
    }
}


function countTheScore(payments, targetAmount, sum) {
    amount = payments.length;
    difference = sum - targetAmount;
    return amount + difference;
}

console.log(solutions.sort((a, b) => a.score - b.score));
// const sortedSolutions = solutions.sort((a, b) => a.score - b.score);
// const bestSolutions = [];
// for (var j=0; j < 5; j++) {
//     bestSolutions.push(sortedSolutions[j]);
// }
// console.log(bestSolutions);

＆＃13;

所有解决方案的输出从最佳到最差排序：

[ { sum: 13, score: 4, payments: [ 5, 5, 2, 1 ] },
  { sum: 15, score: 5, payments: [ 5, 5, 5 ] },
  { sum: 14, score: 5, payments: [ 5, 5, 2, 2 ] },
  { sum: 13, score: 5, payments: [ 5, 5, 1, 1, 1 ] },
  { sum: 13, score: 5, payments: [ 5, 2, 2, 2, 2 ] },
  { sum: 13, score: 6, payments: [ 5, 2, 2, 2, 1, 1 ] },
  { sum: 13, score: 7, payments: [ 5, 2, 2, 1, 1, 1, 1 ] },
  { sum: 13, score: 8, payments: [ 5, 2, 1, 1, 1, 1, 1, 1 ] },
  { sum: 13, score: 9, payments: [ 5, 1, 1, 1, 1, 1, 1, 1, 1 ] } ]

（分数越低越好）

根据您的原始要求，这将是最佳5的输出：

[ { sum: 90, score: 6, payments: [ 50, 20, 20 ] },
  { sum: 90, score: 7, payments: [ 50, 20, 10, 10 ] },
  { sum: 87, score: 7, payments: [ 50, 20, 10, 2, 2, 2, 1 ] },
  { sum: 87, score: 8, payments: [ 50, 10, 10, 10, 2, 2, 2, 1 ] },
  { sum: 90, score: 8, payments: [ 50, 10, 10, 10, 10 ] } ]

btw：如果我将分数功能更改为

function countTheScore(payments, targetAmount, sum) {
    amount = payments.length*3;
    difference = sum - targetAmount;
    return amount + difference;
}

然后输出可能更准确到您需要的

[ { sum: 90, score: 12, payments: [ 50, 20, 20 ] },
  { sum: 90, score: 15, payments: [ 50, 20, 10, 10 ] },
  { sum: 100, score: 16, payments: [ 100 ] },
  { sum: 90, score: 18, payments: [ 50, 10, 10, 10, 10 ] },
  { sum: 100, score: 19, payments: [ 50, 50 ] } ]

您可以使用这些指标进行更多操作，并找出最适合您的输出。

Algo支付设备金额建议现金支付

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