错误地问或我是傻瓜？

Question

Paul Jungwirth 对codinghorror.com进行了博客评论,其中包括一些编程任务:

你有这个数字123456789.在每个数字之间,您必须插入任何内容,加号或乘法符号,以便生成的表达式等于2001.编写一个打印所有解决方案的程序.(那里有两个.)

无聊,我想,我有一个去,但如果我能得到2001年的结果,我会被诅咒.我认为下面的代码是合理的,我认为2001年有零解决方案.根据我的说法.代码,2002年有两个解决方案.我是对还是错了？

/**
 * 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";
    }
}

Answer 1

在你链接的同一页面的下方.... =)

"Paul Jungwirth写道:

你有这个数字123456789.在每个数字之间,您必须插入任何内容,加号或乘法符号,以便生成的表达式等于2001.编写一个打印所有解决方案的程序.(那里有两个.)

我想你的意思是2002年,而不是2001年.:)

(只是纠正像我这样的任何其他人,他们痴迷于尝试解决像这样的小"练习"问题,然后在他们的结果与陈述的答案不符时点击Google.;)该死的,其中一些Perl例子很难看.) "