Programmieren lernen

for a in range(2,20000000):
        if '0' in str(a) or '0' in str(int(10**9/a)): continue
        if (10**9/a).is_integer():
                print(str(a)+' * '+str(int(10**9/a))+' =',str(10**9))
                break

                

Lösung von: rob ert (tub)

inp = 1_000_000_000
trailing_zeros = len(s := str(inp)) - len(s.rstrip("0"))
assert "0" not in str(a := 2**trailing_zeros)
print(f"{a} * {inp//a} = {inp}")
                

Lösung von: Name nicht veröffentlicht

Number.prototype.endsWith = function(num) {
  return this.toString().endsWith(num.toString());
}
Number.prototype.contains0 = function() {
  return this.toString().includes('0');
}

const LIM = Math.ceil(Math.sqrt(1e9));
let result = [],
    i = 2;

console.time();
while (i <= LIM) {
  if (!i.contains0()) {
    let j = 1e9 / i;
    if (Number.isInteger(j) && !j.contains0())
      result.push([i,j]);
  }
  switch (true) {
    case i.endsWith(8): i += 4; break;
    case i.endsWith(2):
    case i.endsWith(6): i += 2; break;
    default: i++;
  }
}
console.timeEnd();
console.log(result);
                

Lösung von: Lisa Salander (Heidi-Klum-Gymnasium Bottrop)

// NET 6.0 | C# 10.x | VS 2022

var n = 1_000_000_000;
var r = n.GetPrimeFactors();
Console.WriteLine($"{n} = {r.GetResult()}");

// Lösung mittels Primfaktorzerlegung -> O(log n)
// Zeit: 10 ms
public static class Extensions {
    public static IEnumerable<int> GetPrimeFactors(this int n) {
        var i = 2;
        while (i <= n) {
            if (n % i == 0) {
                n /= i;
                yield return i;
            }
            else i++;
        }
    }

    public static string GetResult(this IEnumerable<int> lst) => string.Join(" * ", lst.GroupBy(x => x).Select(x => new { b = x.Key, e = x.Count() }).Select(x => $"{ Math.Pow(x.b, x.e) }"));
}
                

Lösung von: Jens Kelm (@JKooP)

// C++ 17
#include <iostream>
#include <vector>
#include <map>
#include <iomanip>

// Zeit: 0.03 ms

std::vector<int> get_prime_factors(int n) {
    std::vector<int>v{};
    auto i{ 2 };
    while (i <= n) {
        if (n % i == 0) {
            n /= i;
            v.push_back(i);
        }
        else ++i;
    }
    return v;
}

std::map<int, int> get_grouped_vector(const std::vector<int>& v) {
    std::map<int, int> m;
    for (const auto& i : v)
        if (m.count(i)) m[i]++;
        else m.emplace(i, 1);
    return m;
}

int main() {
    constexpr auto n{ 1'000'000'000 };
    const auto pf{ get_prime_factors(n) };
    const auto gv{ get_grouped_vector(pf) };
    auto it{ gv.begin() };
    const auto x{ pow(it->first, it->second) };
    const auto y{ pow((++it)->first, it->second) };
    std::cout << std::fixed << std::setprecision(0) << n << " -> " << x << " * "  << y << "\n";
}
                

Lösung von: Jens Kelm (@JKooP)

' VBA
' "Microsoft Scripting Runtime" einbinden

Function GetPrimeFactors(ByVal n As Long)
    Dim arr() As Integer, c As Integer
    Dim i As Long
    i = 2
    Do While i <= n
        If n Mod i = 0 Then
            n = n / i
            ReDim Preserve arr(c)
            arr(c) = i
            c = c + 1
        Else
            i = i + 1
        End If
    Loop
    GetPrimeFactors = arr
End Function

Function GetGroupedArray(arr As Variant) As Scripting.Dictionary
    Dim dic As Scripting.Dictionary
    Dim i As Variant
    Set dic = New Scripting.Dictionary
    For Each i In arr
        If dic.Exists(i) Then
            dic(i) = dic(i) + 1
        Else
            dic.Add i, 1
        End If
    Next
    Set GetGroupedArray = dic
End Function

Sub Main()
    Const NUM As Long = 1000000000
    Dim dic As Scripting.Dictionary
    Dim x As Long, y As Long
    Set dic = GetGroupedArray(GetPrimeFactors(NUM))
    x = dic.Keys(0) ^ dic.Items(0)
    y = dic.Keys(1) ^ dic.Items(1)
    Debug.Print x, y
End Sub

                

Lösung von: Jens Kelm (@JKooP)