I.
    1. d
    2. c
        KIMENET: 010111
        
        f(23)
            f(11)
                f(5)
                    f(2)
                    |   f(1)
                    |   |   f(0)
                    |   |   |   ki 0
                    |   |   []
                    |   |   ki 1
                    |   []
                    |   ki 0
                    []
                    ki 1
                []
                ki 1
            []
            ki 1
            
    3. c
        s: "bac2023",  "ba"
        k: 2
        
        s['c'-'a'] = '\0'
        s[2] = '\0'
        
    4. b (rajzolni kell)
    
    5. a  (5,3), (3,1), (2,4), (4,2)
    
II.

    1.  
        a)  
            x: 6907512     690752              69752         9752         
            p: 1       10           100  1000         10000 
            m: -1,     2            5    7            9      
            c: 2           1        5    7     0      9      6
            
            kimenet: 9752
        
        b)  321  421
        
        c)
        
            #include <iostream>
            using namespace std;

            int main()
            {
                int x;
                cin >> x;

                int p = 1;
                int m = -1;

                while (p <= x) {
                    int c = x/p%10;
                    if (c > m) {
                        m = c;
                        p = p * 10;
                    }
                    else {
                        x = x/(p*10) * p + x%p;
                    }
                }

                if (m >= 0)
                    cout << x << endl;
                else
                    cout << "nul" << endl;

                return 0;
            }

        d)
            beolvas x
            p <- 1
            m <- -1
            
            ha p <= x akkor
            |   ismételd
            |   |   c <- [x/p]%10
            |   |   ha c>m akkor
            |   |   |   m <- c
            |   |   |   p <- p*10
            |   |   |különben
            |   |   |   x <- [x/(p*10)]*p + x%p
            |   |   []
            |   |ameddig p>x
            |   []
            []
            
            ha m >= 0 akkor kiír x
            |különben kiír "nul"
            []
            
II.2.
    lac  mare  ocean  ploaie  rau
           s     s
    
    mare   rau
    ocean  ploaie
    ocean  ploaie  rau
    
II.3.
    struct lalea {
        char denumire[21];
        struct {
            int nrFire;
            int pretFir;
        } stoc;
    } f[10];

III.1.
    #include <iostream>
    using namespace std;
    int abundent1(int n)
    {
        double osztok=0;
        for(int i=1;i<=n;i++){
            if(n%i==0) osztok+=i;
        }
        double arany=osztok/n;
        bool felt=true;
        for(int i=1;i<n && felt==true;i++){
            double s=0;
            for(int j=1;j<=i;j++){
                if(i%j==0) s+=j;
            }
            if(s/i>=arany){
                felt=false;
                break;
            }
        }
        if (felt) return 1;
        else return 0;
    }

    int suma_div(int n)
    {
        int s = 0;

        for(int i=1; i <= n; i++){
            if(n % i == 0)
                s += i;
        }

        return s;
    }

    int abundent(int n)
    {
        double arany_n = (double) suma_div(n) / n;

        bool felt = true;

        for(int i = 1; i < n && felt; i++){
            if((double) suma_div(i) / i >= arany_n){
                felt = false;
            }
        }

        if (felt)
            return 1;
        else
            return 0;
    }



    int main()
    {
        int  n;
        cin >> n;
        for(int i=1; i<=n;i++){
            cout << i << ": " << abundent(i) << endl;
        }

        return 0;
    }

III.2.