用
PHP编写的带有MysqL数据库的Web应用程序.
我有一个系统,可以在分摊成本时为许多人计算不同的成本.例如,人员A购买的东西为10,而人员B,C和D应该分摊成本.
因此,该系统应该记录A人10的正记录和B,C和D的10/3的负记录.
但是,当这样做时;四舍五入后B,C和D均为-3.33.当然这总计不到10个.这个问题的最佳解决方法是什么?一个最佳的解决方案是随机化一个人得到的稍微大一点的成本.
一种可能的解决方案是,如果我只让最后一个人的债务为10 – (A B),但如果四个人分摊成本例如13.34则会出现问题.那么不同的部分将是3.34,3.34,3.34和3.32,而最佳分割将是3.34,3.33,3.33.
有些人可能认为,如果有足够的小数,这只是一个问题,当有大量的行时.但在一个经济的系统中,我认为从一开始就拥有一个故障安全系统是很重要的.它需要是可扩展的,并且不会有任何最小的错误.不公平是好的,只是没有错误.
类似的问题:sum divided values problem (dealing with rounding error)
这似乎工作 –
http://jsfiddle.net/nQakD/.
以jQuery为例,但如果你了解PHP,你应该可以轻松地将它转换为PHP.如果你还需要PHP代码,告诉我,我会为你写的.
$(document).ready(function() { var price = 17.48,people = 4,payment = (price/people).toFixed(2),count=0; var payments = []; for(i = 0; i < people; i++) { payments.push(payment); } if(payment*people != price) { var currentPayment = payment*people; $(payments).each(function() { if(currentPayment < price) { currentPayment = (currentPayment-this).toFixed(2); var newPayment = parseFloat(this)+0.01; payments[count] = newPayment.toFixed(2); currentPayment = parseFloat(currentPayment)+parseFloat(newPayment); } else if(currentPayment > price) { currentPayment = (currentPayment-this).toFixed(2); var newPayment = parseFloat(this)-0.01; payments[count] = newPayment.toFixed(2); currentPayment = parseFloat(currentPayment)+parseFloat(newPayment); } count++; }); } $(payments).each(function() { $("#result").append("<b>"+this+"</b><br/>"); }); });
编辑:
$price = 13.34; $people = 4; $payment = (float)$price/$people; $payment = 0.01 * (int)($payment*100); $count = 0; $payments = Array(); for($i = 0; $i < $people; $i++) { array_push($payments,$payment); } if($payment*$people != $price) { $currentPayment = $payment*$people; foreach($payments as $pay) { if($currentPayment < $price) { $currentPayment = $currentPayment-$pay; $currentPayment = 0.01 * (int)($currentPayment*100); $newPayment = (float)$pay+0.01; $newPayment = 0.01 * (int)($newPayment*100); $payments[$count] = $newPayment; $currentPayment = (float)$currentPayment+$newPayment; } else if($currentPayment > $price) { $currentPayment = $currentPayment-$pay; $currentPayment = 0.01 * (int)($currentPayment*100); $newPayment = (float)$pay-0.01; $newPayment = 0.01 * (int)($newPayment*100); $payments[$count] = $newPayment; $currentPayment = (float)$currentPayment+$newPayment; } $count++; } } foreach($payments as $payed) { echo '<b>'.$payed.'</b><br />'; }
编辑2:
这应该修复js问题 – 上面的http://jsfiddle.net/nQakD/更新代码.
编辑3:
编辑了PHP代码和JS代码,因此它适用于所有示例 – http://jsfiddle.net/nQakD/.
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