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Kedaci

Prime Factors 2

#include <map>
std::map<int, int> prime_factors(int n) {
  auto factors = std::map<int, int>{};

  while (n % 2 == 0) {
    factors[2]++;
    n = n / 2;
  }

  for (auto p = 3; p * p <= n; p += 2) {
    while (n % p == 0) {
      factors[p]++;
      n = n / p;
    }
  }

  if (n > 1) {
    factors[n]++;
    n = n / 1;
  }

  return factors;
}
Kedaci

Prime Factors

#include <vector>

std::vector<int> prime_factors(int n) {
  auto factors = std::vector<int>{};
  if (n == 0) { return factors; }
  while (n % 2 == 0) {
    factors.push_back(2);
    n = n / 2;
  }

  for (auto p = 3; p * p <= n ; p += 2) {
    while (n % p == 0) {
      factors.push_back(p);
      n = n / p;
    }
  }

  if (n > 2) {
    factors.push_back(n);
  }
  return factors;
}
Kedacivs.eurydice5717

Digest word into letters (C++ Translation)

Fundamentals
Strings
Code
Diff
  • #include <string>
#include <fmt/ranges.h>

auto digest(const std::string& param) {
  if (param.empty()) return param; // Oh... should check there !!
  return fmt::format("{}", fmt::join(param, " "));
}
    • #include <string>
    • #include <numeric>
    • #include <fmt/ranges.h>
    • auto digest(const std::string& param) {
    • if (param.empty()) return param; // Oh... should check there !!
    • return std::accumulate(std::next(param.cbegin()),param.cend(),
    • std::string(1,param[0]), // Char initializer
    • [](auto &s, auto c) {
    • return s+std::string{{' ',c}}; // initializer_list string initializer
    • });
    • return fmt::format("{}", fmt::join(param, " "));
    • }
Kedacivs.MariaSvintsova

Return 1!

Code
Diff
  • def should_return_1():
    return (1 << 2) ^ (1 << 2) | 1
    • def should_return_1():
    • return 2 - 1
    • return (1 << 2) ^ (1 << 2) | 1
Kedacivs.Trey2023

Number - 7

Code
Diff
  • class Solution:
    def __init__(self, n):
        self.n = n
        
    def number_se7en(self):
        return self.n // 7 + 1
    • class Solution:
    • def __init__(self, n):
    • self.n = n
    • self.n = n
    • def number_se7en(self):
    • return (self.n - 1) // 7 + 1
    • return self.n // 7 + 1
Kedacivs.KapHullloK

Return 1!

Code
Diff
  • def should_return_1():
    return 0xFF >>  0x7
    • def should_return_1():
    • x = 212312321213125212312321213125 - 212312321213125212312321213124
    • return x
    • return 0xFF >> 0x7
Kedacivs.Kedaci

Return 1!

Code
Diff
  • def should_return_1():
    return (3 + 1 + 1) - (3 + 1 + 0)
    • def should_return_1():
    • return (9999 * 9999 * 9999 * 9999 * 9999 * 9999) ** 0
    • return (3 + 1 + 1) - (3 + 1 + 0)
Kedacivs.seraph776

Return 1!

Code
Diff
  • def should_return_1():
    return (9999 * 9999 * 9999 * 9999 * 9999 * 9999) ** 0
    • def should_return_1():
    • return int(str(int(bin(76)[2:]) & int(bin(7)[2:]))[:1])
    • return (9999 * 9999 * 9999 * 9999 * 9999 * 9999) ** 0
Kedacivs.Kedaci

Return 1000, keeping n immutable

Code
Diff
  • thousand = lambda n = 8: (4 << 4) | (5 << 5) | (6 << 6) | (7 << 7) | n
    • thousand = lambda n = 8: (n << 7) - (n << 1) - n
    • thousand = lambda n = 8: (4 << 4) | (5 << 5) | (6 << 6) | (7 << 7) | n
Kedacivs.monkey_on_a_keyboard_1

Return 100

Code
Diff
  • def return_hundred():
    return (3 << 5) | (1 << 2)
    • def return_hundred():
    • return [int(x) for x in [bin(True+True+True+True)[True+True:]]][False]
    • return (3 << 5) | (1 << 2)
Kedacivs.ramecnt

Return 100

Code
Diff
  • def return_hundred():
    return (9 * 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1)
    • def return_hundred():
    • if 8 == 0:
    • return 100 + 10
    • else:
    • return 100 - 0 + 0 - 0
    • return (1829-1762+21)/100 + 99.12
    • return (9 * 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1)
Kedacivs.Kedaci

Return 1000, keeping n immutable

Code
Diff
  • thousand = lambda n = 8: (n << 7) - (n << 1) - n
    • thousand = lambda n = 8: n * 125
    • thousand = lambda n = 8: (n << 7) - (n << 1) - n
Kedacivs.sreekarthi47

Return 1000, keeping n immutable

Code
Diff
  • thousand = lambda n = 8: n * 125
    • thousand = lambda: 1000
    • thousand = lambda n = 8: n * 125
Kedacivs.seraph776

Return 1000, keeping n immutable

Code
Diff
  • def thousand(n=8):
    return sum(n * x for x in range(15, 40, 5))
    • def thousand(n=8):
    • return [i + (n * 1 + 2) ** 3 for i in range((n * 1 + 2)+5)][0]
    • return sum(n * x for x in range(15, 40, 5))
Kedacivs.seraph776

Return 1000, keeping n immutable

Code
Diff
  • def thousand(n=8):
    return (100 * n) + (20 * n) + (5 * n)
    • def thousand(n=8):
    • return [i + (n * 1 + 2) ** 3 for i in range((n * 1 + 2)+5)][0]
    • return (100 * n) + (20 * n) + (5 * n)
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