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August 23, 2012 09:01
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Project Euler 問題32
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#!/usr/bin/perl | |
use strict; | |
use warnings; | |
use Math::Combinatorics; | |
my @digits = (1,2,3,4,5,6,7,8,9); | |
my %ans = (); | |
#<事前調査その1> | |
#n*mの積は 最小n+m-1桁, 最大n+m桁 になることがわかっている | |
#(1..9)の異なる数字を使った2項間の積の計算式が成立するのは積の値が4桁の時しかない | |
#4桁の積の候補を取得する(順列組み合わせ) | |
my @mul = map { join("", @$_) } (map { permute(@$_) } combine(4, @digits)); | |
foreach my $mul (@mul) | |
{ | |
#print "[$mul]\n"; | |
#残った数字を取得(5個) | |
my @remain = split(//, $mul); | |
@remain = arr_subtract(\@digits, \@remain); | |
#<事前調査その2> | |
#残った5個の数字からn桁とm桁の2項を生成する | |
#(n, m) = (1, 4), (2, 3)の時のみ4桁の数値になることが判明している | |
foreach (1..2) | |
{ | |
my @e1 = combine($_, @remain); #項1を構成する数字のグループ | |
foreach my $_e1 (@e1) | |
{ | |
my @e2 = arr_subtract(\@remain, $_e1); | |
my @_e1 = permute(@$_e1); #項1:順列組み合わせ | |
my @_e2 = permute(@e2); #項2:順列組み合わせ | |
#項1、項2の候補をしらみ潰しに探索 | |
foreach (@_e1) | |
{ | |
my $v1 = join("", @$_); | |
my $v2; | |
#乗数1は入力した値がそのまま出てくるのでパス | |
not ($v1 == 1) or next; | |
foreach (@_e2) | |
{ | |
$v2 = join("", @$_); | |
not ($v1*$v2 == $mul) or $ans{$mul} = 1; | |
} | |
} | |
} | |
} | |
} | |
my $sum = 0; | |
map { $sum += $_ } keys %ans; | |
print "Ans: $sum\n"; | |
exit; | |
# | |
# 配列の残りの要素を取得 | |
# | |
sub arr_subtract | |
{ | |
my $arr1 = shift; | |
my $arr2 = shift; | |
my %d = map { $_ => 1 } @$arr1; | |
map { delete $d{$_} } @$arr2; | |
return (sort keys %d); | |
} | |
# Ans: 45228 |
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遅い