title: Sum Largest Prime Factor Number Less Equal N
signature: sumLargestPrimeFactorNumberLessEqualN :: Integer -> Integer
  sumLargestPrimeFactorNumberLessEqualN n = 
test_code: import Text.Printf
  import Control.Exception (catch, SomeException)
  import Data.Typeable
  import qualified Data.List as List
  import qualified Data.Map as Map
  import qualified Data.Set as Set
  
  -- SOLUTION CODE
  -- ============================================
  
  PLACEHOLDER_CODE_BODY
  
  -- TESTING CODE 
  -- ============================================
  validateSolution :: Integer -> Integer -> Bool
  validateSolution actual expected = do
      actual == expected
  
  
  driver :: Integer -> Integer -> String
  driver n expected = do
      if (validateSolution (PLACEHOLDER_FN_NAME n) expected)
          then "PASSED"
      else "FAILED"
  
  main = do
      printf "TEST-0...%s\n" (driver (6) (15) )
  
      printf "TEST-1...%s\n" (driver (35) (291) )
  
      printf "TEST-2...%s\n" (driver (87) (1554) )
  
      printf "TEST-3...%s\n" (driver (91) (1672) )
  
      printf "TEST-4...%s\n" (driver (63) (875) )
  
      printf "TEST-5...%s\n" (driver (11) (43) )
  
      printf "TEST-6...%s\n" (driver (66) (901) )
  
      printf "TEST-7...%s\n" (driver (17) (90) )
  
      printf "TEST-8...%s\n" (driver (92) (1695) )
  
      printf "TEST-9...%s\n" (driver (81) (1334) )
  
source_py: def sum_largest_prime_factor_number_less_equal_n(n):
      prime = ([0] * (n + 1))
      sum = 0
      max = int((n / 2))
      for p in range(2, (max + 1)):
          if (prime[p] == 0):
              for i in range((p * 2), (n + 1), p):
                  prime[i] = p
      for p in range(2, (n + 1)):
          if prime[p]:
              sum += prime[p]
          else:
              sum += p
      return sum
source_cpp: int sum_largest_prime_factor_number_less_equal_n(int n) {
    int prime [ n + 1 ], sum = 0;
    memset ( prime, 0, sizeof ( prime ) );
    int max = n / 2;
    for ( int p = 2;
    p <= max;
    p ++ ) {
      if ( prime [ p ] == 0 ) {
        for ( int i = p * 2;
        i <= n;
        i += p ) prime [ i ] = p;
      }
    }
    for ( int p = 2;
    p <= n;
    p ++ ) {
      if ( prime [ p ] ) sum += prime [ p ];
      else sum += p;
    }
    return sum;
  }