CodeIQ：ボーリングの最高点、最低点は？

問題の概要

私のプログラム

Rubyで解答しています。

#!/usr/bin/ruby

$Min = 1000
$Max = 0

def calcMin(ptn)
	ret = 0

	for i in 0...10
		if ptn[i] == "a" then
			if ptn[i+1] == "a" then
				if ptn[i+2] == "a" then ret += 30
				else ret += 20
				end
			elsif ptn[i+1] == "b" then ret += 20
			else ret += 10
			end
		elsif ptn[i] == "b" then
			if ptn[i+1] == "a" then ret += 20
			else ret += 10
			end
		else
			ret += 0
		end
	end

	return ret
end

def calcMax(ptn)
	ret = 0

	for i in 0...10
		if ptn[i] == "a" then
			if ptn[i+1] == "a" then
				if ptn[i+2] == "a" then ret += 30
				else ret += 29
				end
			elsif ptn[i+1] == "b" then ret += 20
			else ret += 19
			end
		elsif ptn[i] == "b" then
			if ptn[i+1] == "a" then ret += 20
			else ret += 19
			end
		else
			ret += 9
		end
	end

	return ret
end

def solve(f, a, b, c, ptn)
	if f == 10 then
		# 10フレーム目のパターン
		#    a*3			XXX : OK
		#    a*2 b*1		XX/ : NG	X/X : NG	/XX : NG
		#    a*2 c*1		XXn : OK	XnX : NG	nXX : NG
		#    a*1 b*2		X// : NG	/X/ : NG	//X : NG
		#    a*1 b*1 c*1	X/	: OK	Xn/ : NG	/X  : OK	/nX : NG	nX/ : NG	n/X : NG
		#    a*1 c*2		Xn	: OK	nXn : NG	nnX : NG
		#    b*3			/// : NG
		#    b*2 c*1		//n : NG	/n/ : NG	n// : NG
		#    b*1 c*2		/n	: OK	n/	: NG
		#    c*3			n	: OK
		ptn1 = nil
		ptn2 = nil
		if (a == 3) then ptn1 = ptn + "aaa"
		elsif (a == 2) && (c == 1) then ptn1 = ptn + "aac"
		elsif (a == 1) && (b == 1) && (c == 1) then
			ptn1 = ptn + "ab"
			ptn2 = ptn + "ba"
		elsif (a == 1) && (c == 2) then ptn1 = ptn + "ac"
		elsif (b == 1) && (c == 2) then ptn1 = ptn + "bc"
		elsif (c == 3) then ptn1 = ptn + "c"
		end

		if ptn1 != nil then
			mn = calcMin(ptn1)
			$Min = mn if $Min > mn

			mx = calcMax(ptn1)
			$Max = mx if $Max < mx
		end

		if ptn2 != nil then
			mn = calcMin(ptn2)
			$Min = mn if $Min > mn

			mx = calcMax(ptn2)
			$Max = mx if $Max < mx
		end

		return
	end

	solve(f+1, a-1, b, c, ptn + "a") if a > 0
	solve(f+1, a, b-1, c, ptn + "b") if b > 0
	solve(f+1, a, b, c-1, ptn + "c") if c > 0
end

# main
while line = gets
	line.strip!
	next if line.empty?

	a, b = line.split.map{|v| v.to_i}
	solve(1, a, b, 12-(a+b), "")
	p $Max - $Min
end

解説

雑感

calcMin()とcalcMax()は一つの関数にまとめられました。
もっとうまい方法があるかもしれませんが、まずまずの方法と思います。