title: Find Sum Odd Factors Number
signature: pub fn find_sum_odd_factors_number(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(
        20, 
        6
      ));
      
      println!("TEST-1...{}", driver(
        6, 
        4
      ));
      
      println!("TEST-2...{}", driver(
        39, 
        56
      ));
      
      println!("TEST-3...{}", driver(
        80, 
        6
      ));
      
      println!("TEST-4...{}", driver(
        88, 
        12
      ));
      
      println!("TEST-5...{}", driver(
        7, 
        8
      ));
      
      println!("TEST-6...{}", driver(
        16, 
        1
      ));
      
      println!("TEST-7...{}", driver(
        27, 
        40
      ));
      
      println!("TEST-8...{}", driver(
        83, 
        84
      ));
      
      println!("TEST-9...{}", driver(
        6, 
        4
      ));
      
  }
source_py: def find_sum_odd_factors_number(n):
      res = 1
      while ((n % 2) == 0):
          n = (n // 2)
      for i in range(3, int((math.sqrt(n) + 1))):
          count = 0
          curr_sum = 1
          curr_term = 1
          while ((n % i) == 0):
              count += 1
              n = (n // i)
              curr_term *= i
              curr_sum += curr_term
          res *= curr_sum
      if (n >= 2):
          res *= (1 + n)
      return res
source_cpp: int find_sum_odd_factors_number(int n) {
    int res = 1;
    while ( n % 2 == 0 ) n = n / 2;
    for ( int i = 3;
    i <= sqrt ( n );
    i ++ ) {
      int count = 0, curr_sum = 1;
      int curr_term = 1;
      while ( n % i == 0 ) {
        count ++;
        n = n / i;
        curr_term *= i;
        curr_sum += curr_term;
      }
      res *= curr_sum;
    }
    if ( n >= 2 ) res *= ( 1 + n );
    return res;
  }