~cpp find = [n+1|n <- [1..], (n*(n+1)/2) <= 3000 <= ((n+1)*(n+2)/2)]
=> n 한 , n 합 n+1 합 3000
-> 파 , . (<=) 함수가 어떤 전달인자를 받는지 알아보세요.
while haskell while ?
-> while 함 ..
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Haskell 하 ?ㅋㅋ 합/택2 한 .
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~cpp #include <iostream> using std::cout; using std::endl; int main() { int sum = 0, integer = 1; while(sum <= 3000) { sum = sum + integer; integer++; } cout << "The smallest 'n' for making the number what we want" << endl; cout << "-->" << integer << endl; cout << "Total sum is " << sum << endl; return 0; }
(|) 형 .
~cpp little_sum (x:xs) y |y <= 0 = 0 |otherwise = x + little_sum xs (y-x)