523066680

题目好，长见识（不好意思占楼了  

my @list = qw/
1509231988042396?9
3700001989091894?0
4313021981042877?1
5301021920050801?X
6110251990062874?0/;

my @num = qw(7 9 10 5 8 4 2 1 6 3 7 9 10 5 8 4 2);
my %map;
my $n = 0;
grep { $map{ $_ } = $n++ } qw(1 0 X 9 8 7 6 5 4 3 2);

grep { print func($_) , "\n" } @list;

sub func
{
	my $s = shift;
	my $sum = 0;
	my $last = $map{ substr($s, -1, 1) };
	my $code;

	grep { $sum += substr($s, $_, 1) * $num[$_] } ( 0 .. 15 );
	grep { $code = $_ if ( ($_ * 2 + $sum) % 11 == $last) } ( 0..9 );

	$s =~s/\?/$code/;
	return $s;
}
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根据 happy 的提醒，加一个补充，通过判断和计算求出缺失的数字而非遍历

my @list = qw/
1509231988042396?9
3700001989091894?0
4313021981042877?1
5301021920050801?X
6110251990062874?0/;

my $n = 0;
my %map;
my @num = qw(7 9 10 5 8 4 2 1 6 3 7 9 10 5 8 4 2);
grep { $map{ $_ } = $n++ } qw(1 0 X 9 8 7 6 5 4 3 2);
grep { print func($_) , "\n" } @list;

sub func
{
    my $s = shift;
    my $sum = 0;
    my $last = $map{ substr($s, -1, 1) };
    my $code;

    grep { $sum += substr($s, $_, 1) * $num[$_] } ( 0 .. 15 );
    $test = $sum % 11;
    $code = ( $last - $test ) / 2;
    $code += 11/2 if ( $code < 0 );
    $code += 11/2 if ( int($code) != $code );

    $s =~s/\?/$code/;
    return $s;
}
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523066680

回复 4# happy886rr


    想过， SUM1 + 2*n = LastNumber，通过 SUM1的余数 和 LastNumber映射的余数进行计算，可以得到 n ，不过我猜我用的方案可能不够直接，
需要做 if 判断， 所以还是先粗暴循环了（一般都是先贴，后面再改 ）