Programmieren lernen

def nzahl(basis,n ):
    jj=0
    z=basis
    s=''
    while (jj < n):
        s =str(z)
        if s.startswith(str(basis)):
            jj+=1

        z+=1

    return z-1
    
print nzahl(253, 12)
                

function getNthNumberStartingWithX(nth, x) {
  let i = x, count = 0;
   x = x.toString();
   while (count < nth) {
     if (i.toString().indexOf(x) == 0) count++;
     i++;
   }
   return i;
}

console.log(getNthNumberStartingWithX(1000, 9876));
                

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

// NET Core 3.x
// Lösung durch Berechnen des Subtrahenden

using System;

namespace CS_MDL_CORE_Zahlenspiel
{
    class Program
    {
        static void Main(string[] args)
        {
            Console.WriteLine(GetNumber(253, 12));
            Console.WriteLine(GetNumber(253, 112));
            Console.WriteLine(GetNumber(9876, 1000));
        }

        static int GetNumber(int value, int nth)
        {
            var lengthZeros = (int)Math.Log10(nth);
            var leadingZeros = new string('0', lengthZeros);
            var subtrahend = 1;
            var length = 0;
            var exponent = 0;

            if (nth == 0 || nth == 1) return value;

            while (true) // Subtrahenden ermitteln: 2, 12, 112 ...
            {
                var power = (int)Math.Pow(10, exponent);
                if (nth >= subtrahend + power)
                {
                    subtrahend += power;
                    length++;
                }
                else break;
                exponent++;
            }

            return int.Parse(value + (leadingZeros + (nth - subtrahend))[^length..]);
        }
    }
}
                

Lösung von: Jens Kelm (@JKooP)

#include <iostream>
#include <string>
#include <vector>

struct Pair { int val, nth; };

const auto get_number(const Pair& p) {
    const auto leading_zeros{ std::string(long long(log10(p.nth)), '0') };
    auto subtrahend{ 1 }, exponent{ 0 };

    if (p.nth == 0 || p.nth == 1) return p.val;

    while (1) { // Subtrahenden ermitteln: 2, 12, 112, 1112, ... (Zusatzaufgabe)
        const auto power{ int(pow(10, exponent++)) };
        if (p.nth >= subtrahend + power)
            subtrahend += power;
        else break;
    }

    const auto num{ leading_zeros + std::to_string((p.nth - subtrahend)) };
    return std::stoi(std::to_string(p.val) + num.substr(num.length() - (exponent-1ll)));
}

int main() {
    const std::vector<Pair>pairs{ {253, 12}, {253, 112}, {9'876, 1'000}, {9'876, 105'999} };
    for(const auto& pair : pairs)
        std::cout << get_number(pair) << "\n";

    /*  25300
        253000
        9876888
        987694887 (0,04 ms)
    */
}
                

Lösung von: Jens Kelm (@JKooP)

' VBA

Type Pair
    val As Integer
    nth As Integer
End Type

Function GetNumber(pair_ As Pair) As Long
    Set wsf_ = WorksheetFunction
    Dim subtrahend_ As Integer, exponent_ As Integer
    Dim power_ As Long
    Dim leadingZeros_ As String
    
    subtrahend_ = 1
    leadingZeros_ = String(CInt(wsf_.Log10(pair_.nth)), "0")
    
    If pair_.nth = 0 Or pair_.nth = 1 Then
        GetNumber = pair_.val
        Exit Function
    End If
    
    Do While True
        power_ = WorksheetFunction.Power(10, exponent_)
        If pair_.nth > (subtrahend + power_) Then
            subtrahend_ = subtrahend_ + power_
            exponent_ = exponent_ + 1
        Else
            Exit Do
        End If
    Loop
    GetNumber = CLng(CStr(pair_.val) & right(leadingZeros_ & CStr((pair_.nth - subtrahend_)), exponent_))
End Function

Sub Main()
    Dim pair_ As Pair
    pair_.val = 9876
    pair_.nth = 1000
    Debug.Print GetNumber(pair_)
End Sub

                

Lösung von: Jens Kelm (@JKooP)