将 x 数量的金币分配给 n 个盗贼

Question

假设有 10 个小偷和 100 个金币要分配，分配模式是这样的：

Thief 1 gets 1 coin.
Thief 2 gets 2 coins.
Thief 3 gets 3 coins.... and so on upto 10.

当所有小偷都收到硬币后，总硬币将是(1+2+3+... = 55 coins)。现在剩下的硬币是 45。现在，我怎样才能从小偷开始重新分配，但使用最后一个递增的值，而不是再次从 1 个硬币开始？就像第二轮中的第一个小偷应该得到 11 个硬币而不是 1 个，这应该一直持续到所有硬币都分配完并且剩下的硬币为 0。如果最后一个小偷必须得到 7 个硬币但可用的硬币是 3，那么他应该得到 3 个硬币并且分发应该结束了。

我试过这个...

$thieves = 10;
$goldCoins = 100;

$thiefArr = range(1, $thieves);
$assgnCoins = 0;

foreach($thiefArr as $key => $value){

  for($i=1; $i<=$goldCoins; $i++){

    $assgnCoins = $value; // This assigns first round of coins but how to redistribute it again I have no idea. 

  }
  echo "Thief ".$value." will have ".$assgnCoins." gold coins. <br><br>";
}

Answer 1

我不确定我是否正确理解了您的问题，但这是我的看法，希望对您有所帮助：

<?php
$numt = 10; //Total number of thieves
$totalCoins = 100; //Total coins to redistribute
$data = []; //Data array (will contains the index, which is the "thief" number, and the value of a specific index is the total amount of coins for that specific thief

//Loop init: $coins is the amount of coins given to a thief in a specific distribution round
//Loop condition: We loop until there are still coins to redistribute
//Loop statement: We add 1 coin to each redistribution round
// (I start from 0 and use $coins + 1 because I do modulus on $numt ($coins % $numt) which will give me a number from 0 to $numt - 1 (Hint/fact: array indexes start from 0 and not 1)
for ($coins = 0; $totalCoins > 0; $coins++)
{
    //This will always give me a number between 0 and $numt - 1 (so that we know which thief's turn is)
    //(If this is not clear, you can print out $coins and $thiefIndex then you'll see/understand why I do this)
    $thiefIndex = ($coins % $numt);

    //Because we did not initialize $data values, we check if this specific thief's coins amount has been initialized
    if (!isset($data[$thiefIndex]))
    {
        $data[$thiefIndex] = 0; //Every thief starts with 0 coins
    }

    //Because we started from 0, we need to add 1 (this is the amount of coins that this thief ($thiefIndex) will get in this redistribution round
    $coinsToGive = ($coins + 1);

    //If there's not enough coins left, we just give the total coins remaining
    if ($totalCoins < $coinsToGive)
    {
        $coinsToGive = $totalCoins;
    }

    //Here we sum up all coins that a thief receives in a specific redistribution round
    $data[$thiefIndex] += $coinsToGive;

    //We need to subtract the given coins to the total conins that we redistribute (this makes the loop break when it reaches 0 (see the loop condition part)
    $totalCoins -= $coinsToGive;
}

//Use data/print in your case
foreach ($data as $idx => $tot)
{
    echo "Thief ".($idx + 1)." will have ".$tot." gold coins. <br />";
}

编辑：我想你在我写答案的时候编辑了这个问题。无论如何，如果您想查看每个小偷在每一轮重新分配中获得了多少硬币，您可以在第一个循环中（最后）添加一个回声。

如果您是编程新手，您可能不知道 += 或 -= 的含义：

$foo += $bar;

等同于：

$foo = $foo + $bar;

这可以应用于每个算术运算符，如 +、-、*、/（甚至其他符号，如 &= 和 |= 等......但这些将是按位运算）

有关运营商的信息可以在 PHP 网站上在线找到。

Answer 2

<?php

$thieves = range(1, 10);
$availableCoins = 100;
$assignCoins = 0;

while($availableCoins > 0) {
    foreach($thieves as $thief) {
        // increase the coins to assign
        $assignCoins++;
        
        // check if assign coins are greater than the availableCoins
        if ($assignCoins > $availableCoins) {
            $assignCoins = $availableCoins;
        }

        echo "Thief ".$thief." will have ".$assignCoins." gold coins. <br><br>";

        // substract assignedCoins from availableCoins
        $availableCoins -= $assignCoins;
        // break the loop if no coins left
        if ($availableCoins <= 0) {
            break;
        }
    }
}

Answer 3

我现在已经编辑了@julian S 的答案，给出了作业应该给出的结果：

我还添加了更多的 HTMl 输入来演示数组的结构、正在运行的操作以及以可读形式生成的数组。

简而言之：

这里的范围有误。
因为它会用 1,2,3,4,5,6,7,8,9 填充一个数组，这已经是一个数字数组，所以成员 1 的值是 1 ??无用。不！ 所以我们用初始值 = - 创建了一个 10thives weach 数组，因为小偷最初有 = 钱。

我们打印这个数组并添加 pre 让它看起来更好一点:-)

然后我们有两个循环 - 外部 while 循环具有查看是否还有钱的功能。如果还有moey，那就再做一轮。 这是原始海报中缺少的逻辑，因为他确实知道他需要为 10 个小偷设置一个循环。

然后是内部循环

只需从 1 数到小偷的数量。所以每个小偷都会得到它的份额。 $a += $b 是 $a=$a+$b 的简写形式

将 currentcoins 添加到当前数组成员

如果 monex 在你到达 10 个小偷的末尾之前就消失了，则添加一个 break 以离开 for 循环。

最后只打印带有盗贼财富的数组。

0-9 作为数组以 0 开头。

<?php

$thieves = 10;
$availableCoins = 100;
$assignCoins = 0;
$array_thieves= array_fill(0,10,0);
echo "<h2>Initial array with wealth of each thief</h2> ";
echo "<pre>";
print_r($array_thieves)."<br>";
echo"</pre>";

while($availableCoins > 0) {   //inerate as long as there are coins - outer loop
    for($i=0; $i<$thieves; $i++) {     //execute loop once per thieve so 10 x per round -inner loop
        // increase the coins to assign
        $assignCoins++;   //we have started with 0 and each time we add 1

        // check if assign coins are greater than the availableCoins
        //if the number of available coins is smaller then just assign all available coins
        if ($assignCoins > $availableCoins) {
            $assignCoins = $availableCoins;
        }
        $array_thieves[$i] += $assignCoins;   //add the coins to the thied (= array member) with the current number $i  [0-9]

        echo "Thief ". $i ." gets now additional ".$assignCoins." gold coins. He has now a total of: ". $array_thieves[$i] ." <br>";

        // substract assignedCoins from availableCoins
        $availableCoins -= $assignCoins;
        // break the FOR loop if no coins left - otherwise the inner loop will continue
        if ($availableCoins <= 0) {
            break;
        }
    }
}

echo"<h2>Resulting array with all thieves and their wealth</h2>";
echo"<pre>";
print_r($array_thieves);
echo"</pre>";