Paul Jungwirth在codinghorror.com上有一篇博客文章评论,其中包括一些编程任务:
我很无聊,我想可以,但是如果我能得到2001年的结果,我该死的。我认为下面的代码是正确的,我认为2001年的解决方案为零。代码中有2002年的两种解决方案。我是对还是错?
/**
* Take the numbers 123456789 and form expressions by inserting one of ''
* (empty string), '+' or '*' between each number.
* Find (2) solutions such that the expression evaluates to the number 2001
*/
$input = array(1,2,3,4,5,6,7,8,9);
// an array of strings representing 8 digit, base 3 numbers
$ops = array();
$numOps = sizeof($input)-1; // always 8
$mask = str_repeat('0', $numOps); // mask of 8 zeros for padding
// generate the ops array
$limit = pow(3, $numOps) -1;
for ($i = 0; $i <= $limit; $i++) {
$s = (string) $i;
$s = base_convert($s, 10, 3);
$ops[] = substr($mask, 0, $numOps - strlen($s)) . $s;
}
// for each element in the ops array, generate an expression by inserting
// '', '*' or '+' between the numbers in $input. e.g. element 11111111 will
// result in 1+2+3+4+5+6+7+8+9
$limit = sizeof($ops);
$stringResult = null;
$numericResult = null;
for ($i = 0; $i < $limit; $i++) {
$l = $numOps;
$stringResult = '';
$numericResult = 0;
for ($j = 0; $j <= $l; $j++) {
$stringResult .= (string) $input[$j];
switch (substr($ops[$i], $j, 1)) {
case '0':
break;
case '1':
$stringResult .= '+';
break;
case '2':
$stringResult .= '*';
break;
default :
}
}
// evaluate the expression
// split the expression into smaller ones to be added together
$temp = explode('+', $stringResult);
$additionElems = array();
foreach ($temp as $subExpressions)
{
// split each of those into ones to be multiplied together
$multplicationElems = explode('*', $subExpressions);
$working = 1;
foreach ($multplicationElems as $operand) {
$working *= $operand;
}
$additionElems[] = $working;
}
$numericResult = 0;
foreach($additionElems as $operand)
{
$numericResult += $operand;
}
if ($numericResult == 2001) {
echo "{$stringResult}\n";
}
}
最佳答案
在您链接到的同一页面的下方。...=)