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최소정수의합/나휘동

1

...
~cpp
minsum s n
    | sum [1..n] < s = minsum s (n+1)
    | otherwise = (n, sum [1..n])

2

sum ...
~cpp
naturalSum n = n * (n+1) `div` 2
minsum s n
    | naturalSum n < s = minsum s (n+1)
    | otherwise = (n, naturalSum n)
minsum 3000 1

3

합/ , lazy evaluation
~cpp
take 1 [(n,naturalSum n)| n<-[1..], naturalSum n >= s]
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