secondEquation.scm

; bを固定したときの(a,c)の組み合わせを数える
; (count-ac-combs 3) => 3 [(a,c) = (1,1)(1,2)(2,1)]
(define (count-ac-combs b)
	(let* ((t (quotient (* b b) 4))
		   (comb-a<c
		   	(let loop ((a 1) (sum 0))
		    	(if (> (* a (+ a 1)) t)
		        	sum
		            (loop (+ a 1) (+ sum (- (quotient t a) a))))))
		   (comb-a=c (inexact->exact (floor (sqrt t)))))
		(+ (* comb-a<c 2) comb-a=c)))

; 1 <= b <= m を満たす(a,b,c)の組み合わせの総数を数える
; (count-abc-combs 3) => 4 [(a,b,c) = (1,2,1)(1,3,1)(1,3,2)(2,3,1)]
(define (count-abc-combs m)
	(let loop ((b 1) (sum 0))
		(if (> b m)
			sum
			(loop (+ b 1) (+ sum (count-ac-combs b))))))

; == main ==
; 入力された数字をmとして、1 <= b <= m を満たす(a,b,c)の組み合わせの総数を数える
(display (count-abc-combs 3000))