Project Euler 問題28

pe_28.pl

#!/usr/bin/env perl                                                             

use strict;
use warnings;

my $to = shift; #arg[0] 1001
my $sum = 0;
for(my $n=1; $n<=$to; $n+=2){
    $sum += get_sum_corner($n);
}
print "(n = $to) Ans: $sum\n";

sub get_sum_corner{
    my $n = shift;
    my $f;
    if($n == 1)
    {
        $f = 1;
    }
    else
    {
        $f = 4*$n*$n - ($n-1)*6;
    }
    #print "n=$n: sum = $f\n";
    return($f);
}

# (n = 1001) Ans: 669171001

一番外側の4点のみを考える。

n * n (nは奇数）:
(4) = (3)-(n-1) .. (3) = (2)-(n-1)
: :
(1) = n^2 .. (2) = (1)-(n-1)

(1)+(2)+(3)+(4)を式で書くと
f = n^2+(n^2-(n-1))+((n^2-(n-1))-(n-1))+(((n^2-(n-1))-(n-1))-(n-1))

展開
n は奇数
n = 1 の時: f = 1
n > 1 の時: f = 4_n^2-(n-1)_6