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  3. Project Euler #6: Sum square difference
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Project Euler #6: Sum square difference

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  • skbenz
     + 0 comments

    Don't forget to use long's if working in C/C++. The sums can easily exceed 2^31.

  • yonahel
     + 0 comments

    Haskell

    import Control.Applicative
import Control.Monad
import System.IO


sqSumTo :: Int -> Int
sqSumTo n = ((n * (n+1)) `div` 2 ) ^ 2

sumSqTo :: Int -> Int
sumSqTo n = sum [x * x | x <- [1..n]]

main :: IO ()
main = do
    t_temp <- getLine
    let t = read t_temp :: Int
    forM_ [1..t] $ \a0  -> do
        n_temp <- getLine
        let n = read n_temp :: Int
        print (sqSumTo n - sumSqTo n)

  • neiln
     + 0 comments

    Kotlin 100% - Use long for the last testcase.

    private fun sumOfSquares(input: Long, sum: Long): Long{
    return if(input == 0L) sum else  sumOfSquares(input - 1, sum+(input*input))
}

private fun sumOfNumbers(input: Long):Long{
    return (input * (input+1))/2
}

  • sir_bryan_chong
     + 0 comments

    Here's my solution:

    record = {1:1}
def sumOfFirstNNumbersSquared(n):
    if n in record:
        return record[n]
    else:
        record[n] = sumOfFirstNNumbersSquared(n - 1) + n ** 2
        return record[n]


t = int(input().strip())
for a0 in range(t):
    n = int(input().strip())
    
    a = ( n * (n + 1) / 2 ) ** 2
    b = sumOfFirstNNumbersSquared(n)
    print(int( abs(a - b)) )

  • Hannanmuzammil
     + 0 comments
    t = int(input().strip())
for a0 in range(t):
    n = int(input().strip())
    
    sumofsquare = (n*(n+1)//2)**2 
    squareofsum = n*(n+1)*(2*n+1) // 6 
    print(abs(sumofsquare-squareofsum))