Consider a game where 2 players ($A$ and $B$) each has a bunch of cards, and play alternately.
Playing a shot for a player ($A$) consists in taking the first 2 cards from his bunch and comparing them. We distinguish the following two cases:

• Both cards are equal: ($A$) wins the sum of both cards and both cards are removed.
• Both cards are distinct: ($A$) loses the amount of the largest card. He removes it and places the other card (smallest one) down the bunch of his opponent ($B$).

A party is a succession of shots according to the above described model ($A$ plays the first shot). The game ends when a player ($A$ or $B$) has not enough cards to play. The winner is the one who has the most points; he earns the difference between his points and those of his opponent.

We model a player by:
\(\texttt{typedef struct \{ $\cdots$ Cards ; int gain ; \} player ;}\)

1. Complete the previous declaration by the adequate ADT. 
   In the following, you must use only the ADT operations mentioned in 1.
2. Write a function shot that takes as input 2 players and updates them after a shot of the first player. The function returns 0 if the first player cannot play.
3. Write a function party that simulates a party by taking as input 2 players, returns the winning player and the total amount of points he earns (difference between his points and those of his opponent).
4. Without performing calculations, give the complexity of the previous function in terms of number of shots, and justify your answer.

1.	A and B can be considered as queues
2.	
int shot(player *A, player *B) {	
  element eA1, eA2 ; int d;
  if (!Front(A->Cards,&eA1) || !Dequeue(&(A->Cards) || 
  !Front(A->Cards,&eA2) || !Dequeue(&(A->Cards)) return 0 ;	
  if (eA1 == eA2) A->gain+= 2 * eA1 ;	
  else { 	
         A->gain -= max(eA1,eA2) ; 
         Enqueue(&(B->Cards),min(eA1,eA2)) ; 
       }
  return 1 ; 
}

3.
int party(player *A, player *B, char *winner) 	
{ int x ;
  while (shot(A,B) && shot(B,A)) ;	
  if (x=(A->gain-B->gain) > 0) {*winner = 'A' ; return x ;}	
  else if (x<0) {*winner = 'B' ; return -x ;}	
  else { *winner ='N' ; /* Match Nul */ return 0 ;}	
}
4. order of the number of cards