title: Sum Largest Prime Factor Number Less Equal N
signature: function sum_largest_prime_factor_number_less_equal_n(n)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return actual == expected
  end
  
  function driver(n, expected)
      try
  
          if validate_solution(PLACEHOLDER_FN_NAME(n), expected)
              return "PASSED";
          end
          return "FAILED";
      catch exception_obj
          return string(Base.typename(typeof(exception_obj)).wrapper);
      end
  end
  
  function main()
      
      result = driver(6, 15);
      @printf("TEST-0...%s\n",result);
      
      result = driver(35, 291);
      @printf("TEST-1...%s\n",result);
      
      result = driver(87, 1554);
      @printf("TEST-2...%s\n",result);
      
      result = driver(91, 1672);
      @printf("TEST-3...%s\n",result);
      
      result = driver(63, 875);
      @printf("TEST-4...%s\n",result);
      
      result = driver(11, 43);
      @printf("TEST-5...%s\n",result);
      
      result = driver(66, 901);
      @printf("TEST-6...%s\n",result);
      
      result = driver(17, 90);
      @printf("TEST-7...%s\n",result);
      
      result = driver(92, 1695);
      @printf("TEST-8...%s\n",result);
      
      result = driver(81, 1334);
      @printf("TEST-9...%s\n",result);
      
  end
  
  main()
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;
  }