Solution
3重ループをひとつ減らす工夫をするのがコツ?
pythagorean ?= (a, b, c) -> a*a + b*b == c*c for a in [1..1000] for b in [a..1000-a] c = 1000 - a - b if pythagorean? a, b, c console.log a*b*c
順路:CoffeeScriptでProject Euler #10
逆路:CoffeeScriptでProject Euler #8
3重ループをひとつ減らす工夫をするのがコツ?
pythagorean ?= (a, b, c) -> a*a + b*b == c*c for a in [1..1000] for b in [a..1000-a] c = 1000 - a - b if pythagorean? a, b, c console.log a*b*c
順路:CoffeeScriptでProject Euler #10
逆路:CoffeeScriptでProject Euler #8