42 lines
998 B
Ruby
42 lines
998 B
Ruby
descr=%#
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If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.
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{20,48,52}, {24,45,51}, {30,40,50}
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For which value of p ≤ 1000, is the number of solutions maximised?
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#
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best=0
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maxsol=0
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(5..1000).each do |p|
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# find all {a,b,c}
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# such that
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# a<=b and a^2 + b^2 = c^2
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# and a+b+c = p
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nbsol=0
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# sqrtp=Math.sqrt(p).floor
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(1..p).each do |a|
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# a^2 + b^2 = c^2 => b^2 = c^2 - a^2
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# => b = sqrt( c^2 - a^2 )
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#
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# a + b + c = p => b = p - c - a
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#
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(a..p).each do |b|
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tmp=a**2 + b**2
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c = Math.sqrt(tmp)
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break if a + b + c > p
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next if a + b + c < p
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next if c**2 != tmp
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puts "#{p}: (#{a},#{b},#{c})"
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nbsol += 1
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end
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end
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if nbsol > maxsol
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maxsol=nbsol
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best=p
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end
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end
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puts "Best p = #{best} => #{maxsol}"
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