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blatt05.cpp
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blatt05.cpp
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// Loesungsentwurf zu Aufgabe 5, WS 2014/2015
// Programmieren II fuer Mathematiker
#include <iostream>
#include <vector>
#include <string>
#include <stdexcept>
using namespace std;
class dieder
{
private:
int i;
public:
dieder(int i = 0)
: i(i)
{
if (!(0 <= i && i < 10)) throw invalid_argument("value must be between 0 and 9");
}
dieder inv() const
{
static const int inverse[] = { 0, 4, 3, 2, 1, 5, 6, 7, 8, 9 };
return dieder(inverse[i]);
}
dieder operator*(dieder d) const
{
const int & j = d.i;
if (i <= 4)
{
if (j <= 4) return dieder((i + j) % 5);
else return dieder(5 + (i + j) % 5);
}
else
{
if (j <= 4) return dieder(5 + (i - j) % 5);
else return dieder(i - j < 0 ? 5 + i - j : i - j);
}
}
dieder & operator*=(dieder d) { return *this = *this * d; }
bool operator==(dieder d) const { return i == d.i; }
operator int() const { return i; }
};
// Wurde verwendet um die Tabelle mit den Inversen zu bestimmen
void invtable()
{
for (int i = 0; i < 10; ++i)
for (int j = 0; j < 10; ++j)
{
dieder di(i), dj(j);
if (static_cast<int>(di * dj) == 0)
{
cout << static_cast<int>(dj) << ", ";
}
}
cout << endl;
}
/* Berechnet pi^k(i)) fuer k >= 1 */
int perm(int i, int k)
{
static const int pi[] = { 1, 5, 7, 6, 2, 8, 3, 0, 9, 4 };
while (k-- > 0) i = pi[i];
return i;
}
int _main()
{
invtable();
}
int main()
{
/* Eingabe lesen, auf String speichern */
string inp;
cout << "Geben Sie einen Code ein." << endl;
cin >> inp;
vector<int> digits;
digits.reserve(inp.size());
/* Eingabestring in Ziffernkette umwandeln */
for (unsigned k = 0; k < inp.size(); ++k)
{
int j = 0;
switch (inp[k])
{
case '9': case 'z': case 'Z': ++j;
case '8': case 'y': case 'Y': ++j;
case '7': case 'u': case 'U': ++j;
case '6': case 's': case 'S': ++j;
case '5': case 'n': case 'N': ++j;
case '4': case 'l': case 'L': ++j;
case '3': case 'k': case 'K': ++j;
case '2': case 'g': case 'G': ++j;
case '1': case 'd': case 'D': ++j;
case '0': case 'a': case 'A':
digits.push_back(perm(j, k));
break;
default: throw string("Character ivalid");
}
}
/* Produkt in Diedergruppe berechnen */
dieder prod;
for (unsigned i = 0; i < digits.size(); ++i)
{
prod *= dieder(digits[i]);
}
/* Ausgabe in Grossbuchstaben mit angehaengter Pruefziffer */
for (unsigned i = 0; i < inp.size(); ++i)
{
if ('a' <= inp[i] && inp[i] <= 'z')
cout << static_cast<char>(inp[i] - 'a' + 'A');
else
cout << inp[i];
}
cout << prod.inv() << endl;
return 0;
}