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Harman [31]
3 years ago
10

What kind of charger can i use for this

Computers and Technology
2 answers:
marishachu [46]3 years ago
6 0

Answer:

Is that a juul/ vape?

Explanation:

skelet666 [1.2K]3 years ago
4 0
Is that even a reusable? lol
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Write an abstract data type for a queue in Java whose elements are generic elements of type T that have an associated integer pr
Nimfa-mama [501]

Answer:

See explaination

Explanation:

#include<iostream>

const int MAX=20;

class dque

{

private:

int arr[MAX];

int front,rear;

public:

dque();

void addqatbeg(char item);

void addqatend(char item);

int delqatbeg();

int delqatend();

void display();

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};

dque::dque()

{

front=rear=-1;

for(int i=0;i<MAX;i++)

arr[i]=0;

}

void dque::addqatbeg(int item)

{

if(front==0&&rear==MAX-1){cout<<"\nDeque is full"<<endl;return;}

if(front==-1){front=rear=0;arr[front]=item;return;}

if(rear!=MAX-1)

{int c=count();int k=rear+1;

for(int i=1;i<=c;i++){arr[k]==arr[k-1];k--;}

arr[k]=item;front=k;rear++;

}

else{front--;arr[front]=item;}

}

void dque::addqatend(int item)

{

if(front==0&&rear==MAX-1){cout<<"\nDeque is full"<<endl;return;}

if(front==-1){front=rear=0;arr[rear]=item;return;}

if(rear==MAX-1){

int k=front-1;

for(int i=front-1;i<rear;i++){k=i;

if(k==MAX-1)arr[k]=0;

else arr[k]=arr[i+1];

}rear++;

arr[rear]=item;

}

int dque::delqatbeg()

{

if(front==-1){cout<<"\nDeque is empty"<<endl;return0;}

int item=arr[front];

arr[front]=0;

if(front==rear)front=rear=-1;

else

front++;

return item;

}

int dque::delqatend()

{

if(front==-1){cout<<"\nDeque is empty"<<endl;return 0;}

int item=arr[rear];

arr[rear]=0;

rear--;

if(rear==-1)front=-1;return item;

}

void dque::display()

{

cout<<endl<<"front->";

for(int i=0;i<MAX;i++)cout<<""<<arr[i];cout<<"<-rear";

}

int dque::count()

{

int c=0;

for (int i=0;i<MAX;i++){if(arr[i]!=0)c++;}

return c;}

}

7 0
3 years ago
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Que-?? “Nv” no sé qué es eso?
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Answer:

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Generally speaking, what is a “best practice"?
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What’s up, I’m currently in this class, and I was wondering if it’s easy or hard?

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