title: Sum Largest Prime Factor Number Less Equal N
signature: pub fn sum_largest_prime_factor_number_less_equal_n(n: i32) -> i32 {
test_code: use std::collections::HashMap;
  use std::collections::HashSet;
  use std::cmp::*;
  use std::collections::*;
  use std::fmt::*; 
  
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  pub fn validate_solution(actual: i32, expected: i32)->bool{
  return actual==expected
  }
  pub fn driver(n: i32, expected: i32)->String{
      if validate_solution(PLACEHOLDER_FN_NAME(n), expected) {
        return "PASSED".to_string();
      };
      return "FAILED".to_string();
  
  }
  
  fn main() {
      println!("TEST-0...{}", driver(
        6, 
        15
      ));
      
      println!("TEST-1...{}", driver(
        35, 
        291
      ));
      
      println!("TEST-2...{}", driver(
        87, 
        1554
      ));
      
      println!("TEST-3...{}", driver(
        91, 
        1672
      ));
      
      println!("TEST-4...{}", driver(
        63, 
        875
      ));
      
      println!("TEST-5...{}", driver(
        11, 
        43
      ));
      
      println!("TEST-6...{}", driver(
        66, 
        901
      ));
      
      println!("TEST-7...{}", driver(
        17, 
        90
      ));
      
      println!("TEST-8...{}", driver(
        92, 
        1695
      ));
      
      println!("TEST-9...{}", 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;
  }