[题]1-9组成三个三位数，要求比值为1:2:3

[paktc]

题：用1-9组成三个三位数abc, def, ghi, 每个数字只能使用一次，并且要求 abc:def:ghi=1:2:3
例：219,438,657

[rubyish]

192 384 576
219 438 657
273 546 819
327 654 981

Code: [show] | [select all]

#!/usr/bin/perl -w
use 5.010;
use warnings;

sub play {
  number:
    for my $n ( 123 .. 987 / 3 ) {
        my @num = map { split '', $n * $_ } 1 .. 3;
        my %has = ( 0 => 1 );
        for (@num) {
            next number if $has{$_}++;
        }
        say join $", map { $n * $_ } 1 .. 3;
    }

}

play;

Code: [show] | [select all]

sub play2 {
    for my $n ( 123 .. 987 / 3 ) {
        my %num = map { $_, 1 } map { split '', $n * $_ } 1 .. 3;
        next if keys %num != 9;
        next if exists $num{0};

    say join $", map { $n * $_ } 1 .. 3;
}
}
play2;

[523066680]

代码： 全选

my $a,$b,$c,$n;
for my $i ( 123 .. 334 )
{
    ($a, $b, $c) = ($i, $i * 2, $i * 3);
    $n = $a + $b*1000 + $c * 1000000;
    $mask = 0;
    while ( $n >= 1 )
    {
        $x = $n%10;
        $mask |= 1 << $x;
        $n/=10;
    }
    printf "%d %d %d\n", $a, $b, $c if $mask == 0b1111111110;
}

[rubyish]

有意思。有意思。這個。

代码： 全选

$n = $a + $b*1000 + $c * 1000000;

[rubyish]

 
sub play {
    for my $i ( 123 .. 334 ) {
        my ( $a, $b, $c ) = ( $i, $i * 2, $i * 3 );
        my $n    = $a + $b * 1000 + $c * 1000000;
        my $mask = 0;
        while ( $n >= 1 ) {

代码： 全选
        my $x = $n % 10;
        $n /= 10;

        last if $mask & ( 1 << $x );

        $mask |= 1 << $x;
    }

    say "$a $b $c" if $mask == 1022;
}

[rubyish]

代码： 全选

last if $mask & ( 1 << $x );

[523066680]

代码： 全选

last if $mask & ( 1 << $x );

老了思维退化了，你这个改动我还思考了蛮久的，大意是通过 and 操作符提前判断冲突的数字，减少循环次数。