Codif

#include <fstream>
#include <cstring>
#include <algorithm>
#define DMAX 101
using namespace std;

void citire();
void prelucrare();
void gen(int);
void afisare();

struct solutie
{
    char s[DMAX];
    int lg;
};
struct solutie msj[DMAX];
bool criteriu (struct solutie, struct solutie);

int n, lgsol;
char cod[DMAX], sol[DMAX];
bool putere2[DMAX*10];
int exponent[DMAX*10];
char litere[DMAX/8];
int nrmsj;

int main()
{
    citire();
    prelucrare();
    gen(0);
    afisare();
    return 0;
}

void citire()
{
    int i;
    ifstream fin ("codif.in");
    fin>>n;
    fin>>cod;
    fin.close();
    return;
}

void prelucrare()
{
    int i, doi;
    for (i=doi=1; i<=10; ++i, doi*=2)
    {
        putere2[doi]=1;
        exponent[doi]=i-1;
    }
    litere[0]=' ';
    litere[1]='a';
    litere[2]='e';
    litere[3]='i';
    litere[4]='o';
    litere[5]='u';
    litere[6]='m';
    litere[7]='n';
    litere[8]='r';
    litere[9]='s';
    return;
}

void gen(int k)
{
    int i, nr, j;
    if (k>n) return;
    if (k==n)//am terminat codul
    {
        ++nrmsj;
        sol[k]=0;
        strcpy (msj[nrmsj].s, sol);
        msj[nrmsj].lg=lgsol;
        return;
    }

    //generez in continuare
    for (i=1; i<=3; ++i)//am 3 variante de a forma un numar
    {
        nr=0;
        for (j=1; j<=i && j+k-1<n; ++j)
            nr=nr*10+cod[k+j-1]-'0';

        if (putere2[nr])//daca numarul curent este o putere a lui 2, atunci poate reprezenta o litera
        {
            sol[lgsol]=litere[exponent[nr]];
            lgsol++;
            gen (k+i);//generez de unde am ramas
            sol[lgsol]=0;
            lgsol--;//restaurez
        }
    }
    return;
}

void afisare()
{
    int i, j;
    ofstream fout ("codif.out");
    sort (msj+1, msj+nrmsj+1, criteriu);
    fout<<nrmsj<<"\n";
    for (i=1; i<=nrmsj; ++i)
    {
        fout<<msj[i].s;
        fout<<"\n";
    }
    fout.close();
}

bool criteriu (struct solutie a, struct solutie b)
{
    return (a.lg<b.lg || (a.lg==b.lg && strcmp(a.s, b.s)<0));
}