Numéro 36: Formation informatique - problèmes et défis actuels des grandes entreprises

Des questions

On nous donne une feuille Excel qui contient des entiers de 1 à 50, y compris les deux. Cependant, les nombres sont sous forme brouillée et il manque 1 entier. Vous devez écrire un code pour identifier l'entier manquant. Seule la logique est requise.

Le jeu de Tic-Tac-Toe se joue entre deux joueurs et il est au-dessous de l'état après six coups.

Pouvez-vous répondre aux questions suivantes?

• Qui va gagner le match, O ou X?
• Quelle était la sixième marque et à quelle position?

Supposons que les deux joueurs soient suffisamment intelligents.

Tâches

In Game of Nim, two players take turns removing objects from heaps or the pile of stones.
Suppose two players A and B are playing the game. Each is allowed to take only one stone from the pile. The player who picks the last stone of the pile will win the game. Given N the number of stones in the pile, the task is to find the winner, if player A starts the game.

Input:
The input line contains T, denoting the number of test cases. Then T test case follows, a single line of the input containing a positive integer N.

Output:
Print the winner i.e. Player A or Player B for each testcase in a separate line.

Constraints:
1 <= T <= 100
1 <= N <= 1000

Example:
Input:
2
3
15

Sortie: Explication: Testcase 1: le joueur A retire la pierre 1 qui se trouve en haut, puis le joueur B retire la pierre 2 et enfin le joueur A retire la dernière pierre.
Player A
Player A

Given a Queue Q containing N elements. The task is to reverse the Queue. Your task is to complete the function rev(), that reverses the N elements of the queue.

Input:
The first line of input contains an integer T denoting the Test cases. Then T test cases follow. The first line contains N which is the number of elements which will be reversed. Second line contains N space seperated elements.

Output:
For each testcase, in a new line, print the reversed queue.

Your Task:
This is a function problem. You only need to complete the function revqui prend une file d' attente comme paramètre et renvoie la file d'attente inversée . L' impression se fait automatiquement par le code du pilote .

Contraintes: Exemple: Entrée: Explication: Testcase 1: Après avoir inversé les éléments donnés de la file d'attente, la file d'attente résultante sera 6 2 10 1 3 4. Testcase 2: Après avoir inversé les éléments donnés de la file d'attente, la file d'attente résultante sera 1 2 3 4.
1 ≤ T ≤ 100
1 ≤ N ≤ 105
1 ≤ elements of Queue ≤ 105



2
6
4 3 1 10 2 6
4
4 3 2 1
Output :
6 2 10 1 3 4
1 2 3 4

Given N integers, your task is to add these elements to the PriorityQueue. Also, given M integers, the task is to check if element is present in PriorityQueue. If present, print 1 and return the max element of priority queue, and then delete the max element. If not present, print -1.

Input Format:
First line of testcas contains number of testcases T. For each testcase, there will be 4 lines. First line contains N, number of elements to be inserted into the Priority Queue. Second line contains N positive integers separated by space. Third line contains M, fourth line contains M positive integers.

Output Format:
For each testcase, print «1» and max element in newlines if element to be found is present in the PriorityQueue, else print "-1".

Your Task:
Your task is to complete the functions insert(), find(), and delete(), such that it adds, find and delete the elements from the queue respectively.

Constraints:
1 <= T <= 100
1 <= N <= 103
1 <= M <= 103

Example:
Input:
1
8
1 2 3 4 5 2 3 1
5
1 3 2 9 10

Output:
1
5
1
4
1
3
-1
-1

Explanation:
Testcase 1: After inserting given elements, when we find 1, which is present, so we print 1, and then we print the top element of the PriorityQueue which is 5, and delete it. Similarly, when element is not present, just print "-1".

printf("%s\n",(n % 2) ? "Player A" : "Player B");

{
    // add code here.
    stack<long long int> st;
    while(!q.empty())
    {
        int a=q.front();
        st.push(a);
        q.pop();
    }
    while(!st.empty())
    {
        q.push(st.top());
        st.pop();
    }
    return q;
}

static void insert(PriorityQueue<integer> q, int k){
q.add(k);
}

static boolean find(PriorityQueue<integer> q, int k){
return q.contains(k);
}

static int delete(PriorityQueue<integer> q){
int removed = q.remove();
return removed;
}