Project Euler #1: Multiples of 3 and 5

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    t = int(input().strip()) s=[] for a0 in range(t): n = int(input().strip()) s.append(n) u=s[0] v=s[1] s1=[] s2=[] for j in range(1,u): if(j%3==0 or j%5==0): s1.append(j)
    for j in range(1,v): if(j%3==0 or j%5==0): s2.append(j)
    print(sum(s1)) print(sum(s2))

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    public static void mutiples(int n){
            long m1 = (n-1)/3,m2 = (n-1)/5, m3 = (n-1)/15;
            
            long sum1 = 3*((m1*(m1+1)/2)), sum2 = 5*((m2*(m2+1)/2)), sum3 = 15*((m3*(m3+1)/2));
            
            long sum = sum1+sum2-sum3;
            System.out.println(sum);
        }
    
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    import Control.Applicative import Control.Monad import System.IO

    sumTo :: Integer -> Integer sumTo 0 = 0 sumTo n = (n * (n + 1)) div 2

    multSum :: Integer -> Integer multSum 0 = 0; multSum n = total_sum where three_max = n div 3; five_max = n div 5; fifteen_max = n div 15; three_mult = 3 * (sumTo three_max); five_mult = 5 * (sumTo five_max); fifteen_mult = 15 * (sumTo fifteen_max); total_sum = three_mult + five_mult - fifteen_mult

    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 :: Integer let sum = multSum (n - 1) print (sum)

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

    sumTo :: Integer -> Integer sumTo 0 = 0 sumTo n = (n * (n + 1)) div 2

    multSum :: Integer -> Integer multSum 0 = 0; multSum n = total_sum where three_max = n div 3; five_max = n div 5; fifteen_max = n div 15; three_mult = 3 * (sumTo three_max); five_mult = 5 * (sumTo five_max); fifteen_mult = 15 * (sumTo fifteen_max); total_sum = three_mult + five_mult - fifteen_mult

    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 :: Integer let sum = multSum (n - 1) print (sum)

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    for(int a0 = 0; a0 < t; a0++){ int n = in.nextInt(); System.out.println(sumOfMultiples(n-1)); }}

    private static long sumOfMultiples(int n) { return sumDivisibleBy(n, 3) + sumDivisibleBy(n, 5) - sumDivisibleBy(n, 15); } private static long sumDivisibleBy(int n, int k) { int p = n / k; return k * (long)p * (p + 1) / 2; } }