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Compute efficiently pythagore triplets

DricomDragon 5 年之前
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共有 1 個文件被更改,包括 3 次插入1 次删除
  1. 3 1
      tripy.hs

+ 3 - 1
tripy.hs

@@ -1 +1,3 @@
-square n = [x * x | x <- [1..n]]
+square n = [(x, x * x) | x <- [1..n], x * x > 10]
+
+tripy n = [(a, b, c) | a <- [0..(div n 3)], b <- [(a + 1)..(div n 2)], c <- [(b + 1)..n], a^2 + b^2 == c^2, a + b + c == n]