varianta001

S1)
1)d
2)
)
  n = 103456
  n 		z		p		c
  103456	0		1		6
  10345		3		10		5
  1034						4
  103						3
  10		63		100		0
  			963		1000	1
  963


 b) 61 65 67

 c)

  citeste n
  z<-0
  p<-1
  executa
   c<-n%10
   n<-[n/10]
   daca c%3=0 atunci
   	z<-z+p*(9-c)
   	p<-p*10

   scrie z

d)
#include <iostream>
using namespace std;

int main()
{
	int n , z , p ;
	cin >> n;
	z = 0 ; p = 1 ;
	while(n)
	{
	  c = n % 10 ;
	  n /= 10 ;
	  if(c % 3 == 0)
	  {
	    z = z + p *(9-c) ;
	    p *= 10 ;
	  }

	}
	cout << z ;
	return  0;
}
S2)
1) 5,3 c
3)
C.x = (A.x+B.x) / 2 ;
C.y = (A.y+B.y) / 2 ;

5)
#include <iostream>

using namespace std;

int minim(int a , int b)
{
  if(a > b) return b ;
  return a ;
}

int main()
{
int a[12][12] , n , m , i , j ;
cin >> n >> m ;
for(i = 1 ; i <= n ; ++ i)
{
  for(j = 1 ; j  <= m ; ++ j )
     {
       a[i][j] = minim(i , j) ;
       cout << a[i][j] << " ";
     }
     cout << endl ;
}

return 0 ;
}

S3)
1)

2)f(121,1) =


3)#include <iostream>

#include <fstream>

using namespace std;

ifstream fin("bac.txt") ;

int main()
{
  int n , x , counter = 0 ;

  cin >> n ;
  while(fin >> x)
  {
    if(x % n == 0){ cout << x << " "; counter ++ ;}
  }
  if(counter == 0) cout << "NU EXISTA" ;
}

4)
a)
int sub(int v[] , int n , int a)
{
  int cate = 0 ;
  for(int i = 1 ; i <= n ; ++ i)
     if(v[i] < a) cate ++ ;
  return cate ;
}

b)