Problem Statement

A Pythagorean triple is defined as three positive integers

( a , b , c )

{\displaystyle (a,b,c)}

where

a < b < c

{\displaystyle a<b<c}

, and

a

2

b

2

=

c

2

.

{\displaystyle a^{2}+b^{2}=c^{2}.}

They are called primitive triples if

a , b , c

{\displaystyle a,b,c}

are co-prime, that is, if their pairwise greatest common divisors

g c d

( a , b )

g c d

( a , c )

g c d

( b , c )

1

{\displaystyle {\rm {gcd}}(a,b)={\rm {gcd}}(a,c)={\rm {gcd}}(b,c)=1}

. Because of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime (

g c d

( a , b )

1

{\displaystyle {\rm {gcd}}(a,b)=1}

).   Each triple forms the length of the sides of a right triangle, whose perimeter is

P

a + b + c

{\displaystyle P=a+b+c}

.

The task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.

Deal with large values.   Can your program handle a maximum perimeter of 1,000,000?   What about 10,000,000?   100,000,000? Note: the extra credit is not for you to demonstrate how fast your language is compared to others;   you need a proper algorithm to solve them in a timely manner.

Let's start with the solution:

#NoEnv
SetBatchLines, -1
#SingleInstance, Force

; Greatest common divisor, from http://rosettacode.org/wiki/Greatest_common_divisor#AutoHotkey
gcd(a,b) {
	Return b=0 ? Abs(a) : Gcd(b,mod(a,b))
}

count_triples(max) {
	primitives := 0, triples := 0, m := 2
	while m <= (max / 2)**0.5
	{
		n := mod(m, 2) + 1
		,p := 2*m*(m + n)
		, delta := 4*m
		while n < m and p <= max
			gcd(m, n) = 1
				? (primitives++
				, triples += max // p)
				: ""
			, n += 2
			, p += delta
		m++
	}
	Return primitives " primitives out of " triples " triples"
}

Loop, 8
	Msgbox % 10**A_Index ": " count_triples(10**A_Index)