#includeiostream#includestdlib.husing namespace std;class .pdf

•

0 likes•4 views

#include #include using namespace std; class twoStacks { int *arr; int size; int top1, top2; public: twoStacks(int n) // constructor { size = n; arr = new int[n]; top1 = -1; top2 = size; } // Method to push an element x to stack1 void push1(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top1++; arr[top1] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to push an element x to stack2 void push2(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top2--; arr[top2] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to pop an element from first stack int pop1() { if (top1 >= 0 ) { int x = arr[top1]; top1--; return x; } else { cout << \"Stack UnderFlow\"; exit(1); } } // Method to pop an element from second stack int pop2() { if (top2 < size) { int x = arr[top2]; top2++; return x; } else { cout << \"Stack UnderFlow\"; exit(1); } } }; /* Driver program to test twStacks class */ int main() { twoStacks ts(5); ts.push1(5); ts.push2(10); ts.push2(15); ts.push1(11); ts.push2(7); cout << \"Popped element from stack1 is \" << ts.pop1(); ts.push2(40); cout << \"\ Popped element from stack2 is \" << ts.pop2(); return 0; } #include #include using namespace std; class twoStacks { int *arr; int size; int top1, top2; public: twoStacks(int n) // constructor { size = n; arr = new int[n]; top1 = -1; top2 = size; } // Method to push an element x to stack1 void push1(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top1++; arr[top1] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to push an element x to stack2 void push2(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top2--; arr[top2] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to pop an element from first stack int pop1() { if (top1 >= 0 ) { int x = arr[top1]; top1--; return x; } else { cout << \"Stack UnderFlow\"; exit(1); } } // Method to pop an element from second stack int pop2() { if (top2 < size) { int x = arr[top2]; top2++; return x; } else { cout << \"Stack UnderFlow\"; exit(1); } } }; /* Driver program to test twStacks class */ int main() { twoStacks ts(5); ts.push1(5); ts.push2(10); ts.push2(15); ts.push1(11); ts.push2(7); cout << \"Popped element from stack1 is \" << ts.pop1(); ts.push2(40); cout << \"\ Popped element from stack2 is \" << ts.pop2(); return 0; } Solution #include #include using namespace std; class twoStacks { int *arr; int size; int top1, top2; public: twoStacks(int n) // constructor { size = n; arr = new int[n]; top1 = -1; top2 = size; } // Method to push an element x to stack1 void push1(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top1++; arr[top1] = x; } else { cout << \"Stack Overflow\"; exit(1); } } // Method to push an element x to stack2 void push2(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) {.

Education

if (top1 < top2 - 1)
{
top2--;
arr[top2] = x;
}
else
{
cout << "Stack Overflow";
exit(1);
}
}
// Method to pop an element from first stack
int pop1()
{
if (top1 >= 0 )
{
int x = arr[top1];
top1--;
return x;
}
else
{
cout << "Stack UnderFlow";
exit(1);
}
}
// Method to pop an element from second stack
int pop2()
{
if (top2 < size)
{
int x = arr[top2];
top2++;
return x;
}
else

{
cout << "Stack UnderFlow";
exit(1);
}
}
};
/* Driver program to test twStacks class */
int main()
{
twoStacks ts(5);
ts.push1(5);
ts.push2(10);
ts.push2(15);
ts.push1(11);
ts.push2(7);
cout << "Popped element from stack1 is " << ts.pop1();
ts.push2(40);
cout << " Popped element from stack2 is " << ts.pop2();
return 0;
}
#include
#include
using namespace std;
class twoStacks
{
int *arr;
int size;
int top1, top2;
public:
twoStacks(int n) // constructor
{
size = n;
arr = new int[n];
top1 = -1;
top2 = size;
}

// Method to push an element x to stack1
void push1(int x)
{
// There is at least one empty space for new element
if (top1 < top2 - 1)
{
top1++;
arr[top1] = x;
}
else
{
cout << "Stack Overflow";
exit(1);
}
}
// Method to push an element x to stack2
void push2(int x)
{
// There is at least one empty space for new element
if (top1 < top2 - 1)
{
top2--;
arr[top2] = x;
}
else
{
cout << "Stack Overflow";
exit(1);
}
}
// Method to pop an element from first stack
int pop1()
{
if (top1 >= 0 )
{
int x = arr[top1];

top1--;
return x;
}
else
{
cout << "Stack UnderFlow";
exit(1);
}
}
// Method to pop an element from second stack
int pop2()
{
if (top2 < size)
{
int x = arr[top2];
top2++;
return x;
}
else
{
cout << "Stack UnderFlow";
exit(1);
}
}
};
/* Driver program to test twStacks class */
int main()
{
twoStacks ts(5);
ts.push1(5);
ts.push2(10);
ts.push2(15);
ts.push1(11);
ts.push2(7);
cout << "Popped element from stack1 is " << ts.pop1();
ts.push2(40);

cout << " Popped element from stack2 is " << ts.pop2();
return 0;
}
Solution
#include
#include
using namespace std;
class twoStacks
{
int *arr;
int size;
int top1, top2;
public:
twoStacks(int n) // constructor
{
size = n;
arr = new int[n];
top1 = -1;
top2 = size;
}
// Method to push an element x to stack1
void push1(int x)
{
// There is at least one empty space for new element
if (top1 < top2 - 1)
{
top1++;
arr[top1] = x;
}
else
{
cout << "Stack Overflow";
exit(1);
}

}
// Method to push an element x to stack2
void push2(int x)
{
// There is at least one empty space for new element
if (top1 < top2 - 1)
{
top2--;
arr[top2] = x;
}
else
{
cout << "Stack Overflow";
exit(1);
}
}
// Method to pop an element from first stack
int pop1()
{
if (top1 >= 0 )
{
int x = arr[top1];
top1--;
return x;
}
else
{
cout << "Stack UnderFlow";
exit(1);
}
}
// Method to pop an element from second stack
int pop2()
{
if (top2 < size)
{

int x = arr[top2];
top2++;
return x;
}
else
{
cout << "Stack UnderFlow";
exit(1);
}
}
};
/* Driver program to test twStacks class */
int main()
{
twoStacks ts(5);
ts.push1(5);
ts.push2(10);
ts.push2(15);
ts.push1(11);
ts.push2(7);
cout << "Popped element from stack1 is " << ts.pop1();
ts.push2(40);
cout << " Popped element from stack2 is " << ts.pop2();
return 0;
}
#include
#include
using namespace std;
class twoStacks
{
int *arr;
int size;
int top1, top2;
public:
twoStacks(int n) // constructor
{

size = n;
arr = new int[n];
top1 = -1;
top2 = size;
}
// Method to push an element x to stack1
void push1(int x)
{
// There is at least one empty space for new element
if (top1 < top2 - 1)
{
top1++;
arr[top1] = x;
}
else
{
cout << "Stack Overflow";
exit(1);
}
}
// Method to push an element x to stack2
void push2(int x)
{
// There is at least one empty space for new element
if (top1 < top2 - 1)
{
top2--;
arr[top2] = x;
}
else
{
cout << "Stack Overflow";
exit(1);
}
}
// Method to pop an element from first stack

int pop1()
{
if (top1 >= 0 )
{
int x = arr[top1];
top1--;
return x;
}
else
{
cout << "Stack UnderFlow";
exit(1);
}
}
// Method to pop an element from second stack
int pop2()
{
if (top2 < size)
{
int x = arr[top2];
top2++;
return x;
}
else
{
cout << "Stack UnderFlow";
exit(1);
}
}
};
/* Driver program to test twStacks class */
int main()
{
twoStacks ts(5);
ts.push1(5);
ts.push2(10);

ts.push2(15);
ts.push1(11);
ts.push2(7);
cout << "Popped element from stack1 is " << ts.pop1();
ts.push2(40);
cout << " Popped element from stack2 is " << ts.pop2();
return 0;
}

Similar to #includeiostream#includestdlib.husing namespace std;class .pdf

Stack queue

Stack queue

Stack queue

Stack queue

Stack prgs

New folder/jsjf/ArrayStack.classpackage jsjf; publicsynchronizedclass ArrayStack implements StackADT { privatestaticfinal int DEFAULT_CAPACITY = 100; private int top; private Object[] stack; public void ArrayStack(); public void ArrayStack(int); public void push(Object); private void expandCapacity(); public Object pop() throws exceptions.EmptyCollectionException; public Object peek() throws exceptions.EmptyCollectionException; public int size(); public boolean isEmpty(); public String toString(); } New folder/jsjf/ArrayStack.javaNew folder/jsjf/ArrayStack.javapackage jsjf; import jsjf.exceptions.*; import java.util.Arrays; // ------------------------------------------------------- // Author: Yifu Wu // Date: 03/10/16 // Source Name: ArrayStack<T> // Due date: 03/10/16 // Description: /** * An array implementation of a stack in which the bottom of the * stack is fixed at index 0. * * @author Java Foundations * @version 4.0 */ publicclassArrayStack<T>implementsStackADT<T> { privatefinalstaticint DEFAULT_CAPACITY =100; privateint top; private T[] stack; /** * Creates an empty stack using the default capacity. */ publicArrayStack() { this(DEFAULT_CAPACITY); } /** * Creates an empty stack using the specified capacity. * @param initialCapacity the initial size of the array */ publicArrayStack(int initialCapacity) { top =0; stack =(T[])(newObject[initialCapacity]); } /** * Adds the specified element to the top of this stack, expanding * the capacity of the array if necessary. * @param element generic element to be pushed onto stack */ publicvoid push(T element) { if(size()== stack.length) expandCapacity(); stack[top]= element; top++; } /** * Creates a new array to store the contents of this stack with * twice the capacity of the old one. */ privatevoid expandCapacity() { //stack = Arrays.copyOf(stack, stack.length * 2); System.out.println("Expanding stack capacity\n"); T[] temp =(T[])(newObject[2*top]); for(int i=0; i< top; i++) temp[i]= stack[i]; stack = temp; } /** * Removes the element at the top of this stack and returns a * reference to it. * @return element removed from top of stack * @throws EmptyCollectionException if stack is empty */ public T pop()throwsEmptyCollectionException { if(isEmpty()) thrownewEmptyCollectionException("stack"); top--; T result = stack[top]; stack[top]=null; return result; } /** * Returns a reference to the element at the top of this stack. * The element is not removed from the stack. * @return element on top of stack * @throws EmptyCollectionException if stack is empty */ public T peek()throwsEmptyCollectionException { if(isEmpty()) thrownewEmptyCollectionException("stack"); return stack[top-1]; } /** * Returns the number of elements in ...

New folderjsjfArrayStack.classpackage jsjf;publicsynchronize.docx

curwenmichaela

Stack array

rimshailyas1

DATASTRUCTURES PPTS PREPARED BY M V BRAHMANANDA REDDY

Malikireddy Bramhananda Reddy

Ds 2 cycle

Chaitanya Kn

DATA STRUCTURE USING C & C++

mustkeem khan

DSU C&C++ Practical File Diploma

mustkeem khan

#C programming Question 35Implement the functions required for the.docx

ajoy21

Write a program that accepts an arithmetic expression of unsigned integers in postfix notation and builds the arithmetic expression tree that represents that expression. From that tree, the corresponding fully parenthesized infix expression should be displayed and a file should be generated that contains the three address format instructions. This topic is discussed in the week 4 reading in module 2, section II-B. The main class should create the GUI shown below: Pressing the Construct Tree button should cause the tree to be constructed and using that tree, the corresponding infix expression should be displayed and the three address instruction file should be generated. The postfix expression input should not be required to have spaces between every token. Note in the above example that 9+- are not separated by spaces. The above example should produce the following output file containing the three address instructions: Add R0 5 9 Sub R1 3 R0 Mul R2 2 3 Div R3 R1 R2 Inheritance should be used to define the arithmetic expression tree. At a minimum, it should involve three classes: an abstract class for the tree nodes and two derived classes, one for operand nodes and another for operator nodes. Other classes should be included as needed to accomplish good object-oriented design. All instance data must be declared as private. You may assume that the expression is syntactically correct with regard to the order of operators and operands, but you should check for invalid tokens, such as characters that are not valid operators or operands such as 2a, which are not valid integers. If an invalid token is detected a RuntimeException should be thrown and caught by the main class and an appropriate error message should be displayed. Three Adddress Generator Enter Postfix Expression 359 2 3 Construct Tree Infix Expression ((3-(5 9))/ (2* 3 Solution #include #include #include #include #include struct TREE //Structure to represent a tree { char data; struct TREE *right,*left,*root; }; typedef TREE tree; /* Stack Using Linked List */ struct Stack //Structure to represent Stack { struct TREE *data; struct Stack *next,*head,*top; //Pointers to next node,head and top }; typedef struct Stack node; node *Nw; //Global Variable to represent node void initialize(node *&n) { n = new node; n -> next = n -> head = n -> top = NULL; } void create(node *n) { Nw = new node; //Create a new node Nw -> next = NULL; //Initialize next pointer field if(n -> head == NULL) //if first node { n -> head = Nw; //Initialize head n -> top = Nw; //Update top } } void push(node *n,tree *ITEM) { node *temp = n -> head; if(n -> head == NULL) //if First Item is Pushed to Stack { create(n); //Create a Node n -> head -> data = ITEM; //Insert Item to head of List n -> head = Nw; return; //Exit from Function } create(n); //Create a new Node Nw -> data = ITEM; //Insert Item while(temp -> next != NULL) temp = temp -> next; //Go to Last Node temp -> next = Nw; //Point New node n -> top = Nw; //Upda.

Write a program that accepts an arithmetic expression of unsigned in.pdf

JUSTSTYLISH3B2MOHALI

Stack of Data structure

Sheikh Monirul Hasan

Project of data structure

Umme habiba

C++ Searching & Sorting 5. Sort the following list using the selection sort algorithm. Show the list after each iteration of the outerforloop. 36, 55, 17, 35, 63, 85, 12, 48, 3, 66 6. Consider the following list: 5, 18, 21, 10, 55, 20 The first three keys are in order. To move 10 to its proper position using the insertion sort as described in this chapter, exactly how many key comparisons are executed? 7. Consider the following list: 7, 28, 31, 40, 5, 20 The first four keys are in order. To move 5 to its proper position using the insertion sort as described in this chapter, exactly how many key comparisons are executed? 8. Consider the following list: 28, 18, 21, 10, 25, 30, 12, 71, 32, 58, 15 This list is to be sorted using the insertion sort algorithm. Show the resulting list after six passes of the sorting phase – that is, after six iterations of the for loop. 9. Perform the insertion sort algorithm using the following list of keys: 18, 8, 11, 9, 15, 20, 32, 61, 22, 48, 75, 83, 35, 3 Show the list after each iteration. Exactly how many key comparisons are executed to sort this list using insertion sort? 10. a. The performance of bubble sort can be improved if we stop the sorting process as soon as we find that in an iteration, no swapping of elements takes place. Write a function that implements bubble sort algorithm using this fact. b. Using the algorithm that you designed in part (a), find the number of iterations that are needed to sort the list: 65, 14, 52, 43, 75, 25, 80, 90, 95. 11. Suppose that L is a sorted list of 4096 elements. What is the maximum number of comparisons made by binary search to determine whether an item is in L? 12. Suppose that the elements of a list are in descending order, and they need to be put in ascending order. Write a C++ function that takes as input an array of items in descending order and the number of elements in the array. The function must not incorporate any sorting algorithms, that is, no item comparisons should take place. Solution # include # include # include #include #include #include #include // Function related to sorting in class sorting class sorting { int array[50],array1[50],final[100],i,n,m,j; public: // Function to read an array void read(); // Function to read arrays for merge sort void read_mer(); // Function to display an array void display(); // Function to perform bubble sort void bub_sort(); // Function to perform selection sort void Sel_sort(); // Function to perform insertion sort void Ins_sort(); // Function to perform quick sort void Qui_sort(); // Function to perform heap sort void Heap_sort(); // Function to build a heap void heap(int array[], int n); // Function to interchange the value of root node with a // child node in heap sort void below_heap(int array[], int first, int last); // Function to perform merges sort void Mer_sort(); // Function to perform shell sort void Shell_sort(); // Function to split the array into two halves during quick sort void partition(int arra.

C++ Searching & Sorting5. Sort the following list using the select.pdf

Rahul04August

Polymorphism

mohamed sikander

week-14x

KITE www.kitecolleges.com

Datastructures asignment

sreekanth3dce

Add functions push(int n, Deque &dq) and pop(Deque &dq). Functions push(int n, Deque &dq) and pop(Deque &dq) should call functions of the Deque class to simulate a stack's push and pop functions. The Deque class should not be touched . I will use the original unmodified Deque class when testing your code. Demonstrate to convert a base 10 number to base 2. Ask the user to enter a base 10 number, output the equivalent base 2 number. The main() function would create a Deque object and pass it by reference when calling the new push(int n, Deque &dq) and pop(Deque &dq) functions simulating stack operations. keep it simple and to the instructions and code provided please! and thank you! // C++ implementation of De-queue using circular // array #include<iostream> using namespace std; // Maximum size of array or Dequeue #define MAX 100 // A structure to represent a Deque class Deque { int arr[MAX]; int front; int rear; int size; public : Deque(int size) { front = -1; rear = 0; this->size = size; } // Operations on Deque: void insertfront(int key); void insertrear(int key); void deletefront(); void deleterear(); bool isFull(); bool isEmpty(); int getFront(); int getRear(); }; // Checks whether Deque is full or not. bool Deque::isFull() { return ((front == 0 && rear == size-1)|| front == rear+1); } // Checks whether Deque is empty or not. bool Deque::isEmpty () { return (front == -1); } // Inserts an element at front void Deque::insertfront(int key) { // check whether Deque if full or not if (isFull()) { cout << "Overflow\n" << endl; return; } // If queue is initially empty if (front == -1) { front = 0; rear = 0; } // front is at first position of queue else if (front == 0) front = size - 1 ; else // decrement front end by '1' front = front-1; // insert current element into Deque arr[front] = key ; } // function to inset element at rear end // of Deque. void Deque ::insertrear(int key) { if (isFull()) { cout << " Overflow\n " << endl; return; } // If queue is initially empty if (front == -1) { front = 0; rear = 0; } // rear is at last position of queue else if (rear == size-1) rear = 0; // increment rear end by '1' else rear = rear+1; // insert current element into Deque arr[rear] = key ; } // Deletes element at front end of Deque void Deque ::deletefront() { // check whether Deque is empty or not if (isEmpty()) { cout << "Queue Underflow\n" << endl; return ; } // Deque has only one element if (front == rear) { front = -1; rear = -1; } else // back to initial position if (front == size -1) front = 0; else // increment front by '1' to remove current // front value from Deque front = front+1; } // Delete element at rear end of Deque void Deque::deleterear() { if (isEmpty()) { cout << " Underflow\n" << endl ; return ; } // Deque has only one element if (front == rear) { front = -1; rear = -1; } else if (rear == 0) rear = size-1; else rear = rear-1; } // Returns front element of Deque int Deque::getFront() { // check whether Deque is empty or not if (isEmp.

Add functions push(int n- Deque &dq) and pop(Deque &dq)- Functions pus.docx

WilliamZnlMarshallc

Similar to #includeiostream#includestdlib.husing namespace std;class .pdf (20)

Stack queue

Stack prgs

New folderjsjfArrayStack.classpackage jsjf;publicsynchronize.docx

Stack array

DATASTRUCTURES PPTS PREPARED BY M V BRAHMANANDA REDDY

Ds 2 cycle

DATA STRUCTURE USING C & C++

DSU C&C++ Practical File Diploma

#C programming Question 35Implement the functions required for the.docx

Write a program that accepts an arithmetic expression of unsigned in.pdf

Stack of Data structure

Project of data structure

C++ Searching & Sorting5. Sort the following list using the select.pdf

Polymorphism

week-14x

Datastructures asignment

Add functions push(int n- Deque &dq) and pop(Deque &dq)- Functions pus.docx

More from asif1401

the reaction of A to B since its Ea is the highes.pdf

asif1401

the one thing that comes to mind for me is about the diatomic elements H,N,F,O,I,Cl,B they\'re all really the ones that could form a molecule since they are composed of more than one element (all of them are subscript 2, as in H2). most other elements exists as 1. dont confuse this with MOLECULES which are composed of different atoms together. Solution the one thing that comes to mind for me is about the diatomic elements H,N,F,O,I,Cl,B they\'re all really the ones that could form a molecule since they are composed of more than one element (all of them are subscript 2, as in H2). most other elements exists as 1. dont confuse this with MOLECULES which are composed of different atoms together..

the one thing that comes to mind for me is about .pdf

asif1401

(1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen, and other chemicals in the atmosphere to form various acidic compounds. Sunlight increases the rate of most of these reactions. The result is a mild solution of sulfuric acid and nitric acid. Acid rain causes acidification of lakes and streams and contributes to damage of trees at high elevations and many sensitive forest soils. (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution = ( 5.47 x 10 -3 / 46 ) / 1.0 = 1.189 x 10 -4 M According to the balanced equation, 2 mole of NO2 produces 2 moles of HNO3 So the concentration of HNO3 is = 1.189 x 10 -4 M HNO3 ---> H+ + NO3- 1 mole of HNO3 produces 1 mole of H+ So [H+] = 1.189 x 10 -4 M pH = - log [H+] = -log ( 1.189 x 10 -4 ) = 4 - log 1.189 = 3.92 (3) HNO3 + H2O ----> HNO3 (aq) H2SO4 + H2O ---> H2SO4 (aq) HCl + H2O ---> HCl (aq) (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution = ( 5.47 x 10 -3 / 46 ) / 1.0 = 1.189 x 10 -4 M According to the balanced equation, 2 mole of NO2 produces 2 moles of HNO3 So the concentration of HNO3 is = 1.189 x 10 -4 M HNO3 ---> H+ + NO3- 1 mole of HNO3 produces 1 mole of H+ So [H+] = 1.189 x 10 -4 M pH = - log [H+] = -log ( 1.189 x 10 -4 ) = 4 - log 1.189 = 3.92 (3) HNO3 + H2O ----> HNO3 (aq) H2SO4 + H2O ---> H2SO4 (aq) HCl + H2O ---> HCl (aq) Solution (1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen, and other chemicals in the atmosphere to form various acidic compounds. Sunlight increases the rate of most of these reactions. The result is a mild solution of sulfuric acid and nitric acid. Acid rain causes acidification of lakes and streams and contributes to damage of trees at high elevations and many sensitive forest soils. (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Volume of the solution = ( 5.47 x 10 -3 / 46 ) / 1.0 = 1.189 x 10 -4 M According to the balanced equation, 2 mole of NO2 produces 2 moles of HNO3 So the concentration of HNO3 is = 1.189 x 10 -4 M HNO3 ---> H+ + NO3- 1 mole of HNO3 produces 1 mole of H+ So [H+] = 1.189 x 10 -4 M pH = - log [H+] = -log ( 1.189 x 10 -4 ) = 4 - log 1.189 = 3.92 (3) HNO3 + H2O ----> HNO3 (aq) H2SO4 + H2O ---> H2SO4 (aq) HCl + H2O ---> HCl (aq) (2) 2NO2 + H2O ---> 2HNO3 Given mass of NO2 is , m = 5.47 mg = 5.47 x 10 -3 g Since 1 mg = 10 -3 g Volume of the solution , V = 1.0 L Molar mass of NO2 is = 14 + (2x16) = 46 g/mol So Molarity of NO2 , M = ( mass/Molar mass) / Vol.

(1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen.pdf

asif1401

Iodine will replace both Cl and Br It will be SN2.pdf

asif1401

The number of outcomes in the event A is 4 Yes event A is a s.pdf

asif1401

There were very few planetary leftovers in this r.pdf

asif1401

i have no idea. im just seeing if i get karma p.pdf

asif1401

Fe+3 solution is more acidic because it produces H+ ion in the aqueous solution, but Fe+2 is not. The following reaction represents this- [Fe(H2O)6]3+H2O ----------> [Fe(H2O)5OH]+2+ + H3O+(aq) Therefore - a) Fe+3 b) ii Solution Fe+3 solution is more acidic because it produces H+ ion in the aqueous solution, but Fe+2 is not. The following reaction represents this- [Fe(H2O)6]3+H2O ----------> [Fe(H2O)5OH]+2+ + H3O+(aq) Therefore - a) Fe+3 b) ii.

Fe+3 solution is more acidic because it produces .pdf

asif1401

Copper chloride. Copper ions will form precipitated with sulfide, Coppersulfide. For other cations, there are solubile when there form with eithersulfide ions or acetate ions. So, the only way to separate is using copper ions. Solution Copper chloride. Copper ions will form precipitated with sulfide, Coppersulfide. For other cations, there are solubile when there form with eithersulfide ions or acetate ions. So, the only way to separate is using copper ions..

Copper chloride. Copper ions will form precipitat.pdf

asif1401

CCl4 molecule can give IR and Raman spectroscopy. Selection rule of IR spectra: During the vibration dipolemoment change. Selection rule of Raman spectra: During the vibration polarizibility change. CCl4 molecule centrosymmetric molecule, so it can exibit Raman spectroscopy According to mutual exclusion principle, IR inactive modes are raman active Raman inactive modes are IR active modes. Therefore, CCl4 molecule can give multiple peaks. Solution CCl4 molecule can give IR and Raman spectroscopy. Selection rule of IR spectra: During the vibration dipolemoment change. Selection rule of Raman spectra: During the vibration polarizibility change. CCl4 molecule centrosymmetric molecule, so it can exibit Raman spectroscopy According to mutual exclusion principle, IR inactive modes are raman active Raman inactive modes are IR active modes. Therefore, CCl4 molecule can give multiple peaks..

CCl4 molecule can give IR and Raman spectroscopy. Selection r.pdf

asif1401

User interface design processNavigation DesignInput DesignOu.pdf

asif1401

True.it is an example of an independent-measures designSolution.pdf

asif1401

The main class of the tictoe game looks like.public class Main {.pdf

asif1401

the inner electronsSolutionthe inner electrons.pdf

asif1401

Solution The investee\'s net income should be recorded as an increase to the investment account. False Explanation Not all the net income of investee is recorded as an increase to investment account. Only the part of net income proportionate to investor share is recorded as increase in investment account. For example if investor have 30% share in investee company and the net income of investee is 100000 then only 30000 will be recorded as an increase in investment account.

SolutionThe investees net income should be recorded as an increa.pdf

asif1401

Question Answer 1 True The corpus callosum , otherwise called the callosal commissure, is a wide, level heap of neural filaments around 10 cm long underneath the cortex in the eutherian cerebrum at the longitudinal gap. It associates the left and right cerebral sides of the equator and encourages interhemispheric correspondence. It is the biggest white matter structure in the mind, comprising of 200–250 million contralateral axonal projections. The (back) bit of the corpus callosum is known as the splenium; the foremost (front) is known as the genu (or \"knee\"); between the two is the truncus, or \"body\", of the corpus callosum. The part between the body and the splenium is frequently extraordinarily contracted and along these lines alluded to as the \"isthmus\". The platform is the part of the corpus callosum that ventures posteriorly and poorly from the anteriormost genu, as can be seen on the sagittal picture of the cerebrum showed on the privilege. The platform is so named for its likeness to a winged creature\'s mouth. Corpus callosum On either side of the corpus callosum, the filaments transmit in the white matter and go to the different parts of the cerebral cortex; those bending forward from the genu into the frontal flap constitute the forceps foremost, and those bending in reverse into the occipital projection, the forceps back. Between these two sections is the fundamental body of the filaments which constitute the tapetum and expand along the side on either side into the transient projection, and cover in the focal part of the horizontal ventricle. 2 False Ramus communicans (plural rami communicantes) is the Latin expression utilized for a nerve which associates two different nerves, and can be deciphered as \"conveying branch\".When utilized without further definition, it quite often alludes to an imparting branch between a spinal nerve and the thoughtful trunk. All the more particularly, it generally alludes to one of the accompanying : Dark ramus communicans White ramus communicans The dark and white rami communicantes are in charge of passing on autonomic signs, particularly for the thoughtful sensory system. Their distinction in tinge is brought about by contrasts in myelination of the nerve filaments contained inside, i.e. there are more myelinated than unmyelinated filaments in the white rami communicantes while the opposite is valid for the dim rami communicantes. 3 False tentorium cerebelli isolates the occipital flaps of the cerebral sides of the equator from the cerebellum. Its inward, sunken, free fringe adds to the tentorial indent. (See underneath.) The outer, arched fringe encases the transverse sinus where it appends to the dura over within the occiput. Past the \"petrous edge,\" the tentorium is tied down to the foremost and back clinoid forms 4 False myelin sheath(deep) and Neurilemma (superficial) 5 False Cranial nerves are the nerves that develop straightforwardly from the cerebrum (counting the brainstem), a.

QuestionAnswer1TrueThe corpus callosum , otherwise called th.pdf

asif1401

No. of Moles = Given MassMolecular Mass=3.0 x 101398= 3.06101.pdf

asif1401

Let the sample size be n ,Mean of the sample = 27 (same as pop.pdf

asif1401

JeffSolutionJeff.pdf

asif1401

import java.util.Scanner; public class Assignment4 { public static void main(String args[]) { final int LEN = 5; int[] hand = new int[LEN]; Scanner input = new Scanner(System.in); //input the hand System.out.println(\"Enter five numeric cards, no face cards. Use 2-9.\"); for (int i = 0; i < hand.length; i++) { hand[i] = input.nextInt(); } //sort the collection bubbleSortCards(hand); //determine players hand type //flow of evaluation -> checking complex hands first if (containsFullHouse(hand)) { System.out.println(\"Full House!\"); } else if (containsStraight(hand)) { System.out.println(\"Straight!\"); } else if (containsFourOfaKind(hand)) { System.out.println(\"Four of a Kind!\"); } else if (containsThreeOfaKind(hand)) { System.out.println(\"Three of a Kind!\"); } else if (containsTwoPair(hand)) { System.out.println(\"Two Pair!\"); } else if (containsPair(hand)) { System.out.println(\"Pair!\"); } else System.out.println(\"High Card!\"); } //cardInSets is called with hand[] and pair to check if a pair is found. public static boolean containsPair(int hand[]) { int pair = 2; return cardInSets(hand, pair) != NOT_FOUND; } //cardInSets is called with hand[] and pair to check if a pair is found. //Then calls back again with fcard result to check for another pair. public static boolean containsTwoPair(int hand[]) { int pair = 2; int fcard = cardInSets(hand, pair); int scard = cardInSets(hand, pair, fcard); return fcard!=NOT_FOUND && scard!=NOT_FOUND; } //cardInSets is called with hand[] and threeaKind to check if three of a kind has been found. public static boolean containsThreeOfaKind(int hand[]) { int threeaKind = 3; return cardInSets(hand, threeaKind) != NOT_FOUND; } //hasStraight will resolve to true, unless the previous value++ is not equivalent. //The loop ends as soon as hasStaight is false or the end of the hand is reached. public static boolean containsStraight(int hand[]) { boolean hasStraight = true; int i=1; do { hasStraight = (hand[i] == (hand[i-1] + 1)); } while(hasStraight && ++i < hand.length); return hasStraight; } //cardInSets is called with hand[] and threeKinds to check if three of a kind is found. //Then calls back again with pair and tcard result to check for a pair of another card type. public static boolean containsFullHouse(int hand[]) { int threeKinds = 3; int pair = 2; int tcard = cardInSets(hand, threeKinds); int pcard = cardInSets(hand, pair, tcard); return tcard!=NOT_FOUND && pcard!=NOT_FOUND; } //cardInSets is called with hand[] and fouraKind to check if four of a kind is found. public static boolean containsFourOfaKind(int hand[]) { int fouraKind = 4; return cardInSets(hand, fouraKind) != NOT_FOUND; } //When ignoreCard isn\'t needed, it overloads with a sentinel 0. public static int cardInSets(int hand[], int numKinds) { return cardInSets(hand, numKinds, 0); } //The method cardInSets is a generic evaluation that returns if the parameter specified numKinds(like 2 for pair) has been found in the hand. //The ignoreCard d.

import java.util.Scanner;public class Assignment4 { public st.pdf

asif1401

More from asif1401 (20)

the reaction of A to B since its Ea is the highes.pdf

the one thing that comes to mind for me is about .pdf

(1) Acid rain occurs when the gases SO2 ,NOx react with water,oxygen.pdf

Iodine will replace both Cl and Br It will be SN2.pdf

The number of outcomes in the event A is 4 Yes event A is a s.pdf

There were very few planetary leftovers in this r.pdf

i have no idea. im just seeing if i get karma p.pdf

Fe+3 solution is more acidic because it produces .pdf

Copper chloride. Copper ions will form precipitat.pdf

CCl4 molecule can give IR and Raman spectroscopy. Selection r.pdf

User interface design processNavigation DesignInput DesignOu.pdf

True.it is an example of an independent-measures designSolution.pdf

The main class of the tictoe game looks like.public class Main {.pdf

the inner electronsSolutionthe inner electrons.pdf

SolutionThe investees net income should be recorded as an increa.pdf

QuestionAnswer1TrueThe corpus callosum , otherwise called th.pdf

No. of Moles = Given MassMolecular Mass=3.0 x 101398= 3.06101.pdf

Let the sample size be n ,Mean of the sample = 27 (same as pop.pdf

JeffSolutionJeff.pdf

import java.util.Scanner;public class Assignment4 { public st.pdf

Recently uploaded

How to setup Pycharm environment for Odoo 17.pptx

Celine George

On National Teacher Day, meet the 2024-25 Kenan Fellows

Mebane Rash

How to Manage Global Discount in Odoo 17 POS

Celine George

On_Translating_a_Tamil_Poem_by_A_K_Ramanujan.pptx

Pooja Bhuva

This PowerPoint helps students to consider the concept of infinity.

christianmathematics

FSB Advising Checklist - Orientation 2024

Elizabeth Walsh

General Principles of Intellectual Property: Concepts of Intellectual Proper...

Poonam Aher Patil

Fostering Friendships - Enhancing Social Bonds in the Classroom

Pooky Knightsmith

Sensory_Experience_and_Emotional_Resonance_in_Gabriel_Okaras_The_Piano_and_Th...

Pooja Bhuva

Mehran University Newsletter Vol-X, Issue-I, 2024

Mehran University of Engineering & Technology, Jamshoro

Klinik_ Apotek Onlin 085657271886 Solusi Menggugurkan Masalah Kehamilan Anda Jual Obat Aborsi Asli KLINIK ABORSI TERPEECAYA _ Jual Obat Aborsi Cytotec Misoprostol Asli 100% Ampuh Hanya 3 Jam Langsung Gugur || OBAT PENGGUGUR KANDUNGAN AMPUH MANJUR OBAT ABORSI OLINE" APOTIK Jual Obat Cytotec, Gastrul, Gynecoside Asli Ampuh. JUAL ” Obat Aborsi Tuntas | Obat Aborsi Manjur | Obat Aborsi Ampuh | Obat Penggugur Janin | Obat Pencegah Kehamilan | Obat Pelancar Haid | Obat terlambat Bulan | Ciri Obat Aborsi Asli | Obat Telat Bulan | Pil Aborsi Asli | Cara Menggugurkan Konten | Cara Aborsi Tuntas | Harga Obat Aborsi Asli | Pil Aborsi | Jual Obat Aborsi Cytotec | Cara Aborsi Sendiri | Cara Aborsi Usia 1 Bulan | Cara Aborsi Usia 2 Tahun | Cara Aborsi Usia 3 Bulan | Obat Aborsi Usia 4 Bulan | Cara Abrasi Usia 5 Bulan | Cara Menggugurkan Konten | Kandungan Obat Penggugur | Cara Menghitung Usia Konten | Cara Mengatasi Terlambat Bulan | Penjual Obat Aborsi Asli | Obat Aborsi Garansi | Kandungan Obat Peluntur | Obat Telat Datang Bulan | Obat Telat Haid | Obat Aborsi Paling Murah | Klinik Jual Obat Aborsi | Jual Pil Cytotec | Apotik Jual Obat Aborsi | Kandungan Dokter Abrasi | Cara Aborsi Cepat | Jual Obat Aborsi Bergaransi | Jual Obat Cytotec Asli | Obat Aborsi Aman Manjur | Obat Misoprostol Cytotec Asli. "APA ITU ABORSI" “Aborsi Adalah dengan membendung hormon yang di perlukan untuk mempertahankan kehamilan yaitu hormon progesteron, karena hormon ini dibendung, maka jalur kehamilan mulai membuka dan leher rahim menjadi melunak,sehingga mengeluarkan darah yang merupakan tanda bahwa obat telah bekerja || maksimal 1 jam obat diminum || PENJELASAN OBAT ABORSI USIA 1 _7 BULAN Pada usia kandungan ini, pasien akan merasakan sakit yang sedikit tidak berlebihan || sekitar 1 jam ||. namun hanya akan terjadi pada saatdarah keluar merupakan pertanda menstruasi. Hal ini dikarenakan pada usiakandungan 3 bulan,janin sudah terbentuk sebesar kepalan tangan orang dewasa. Cara kerja obat aborsi : JUAL OBAT ABORSI AMPUH dosis 3 bulan secara umum sama dengan cara kerja || DOSIS OBAT ABORSI 2 bulan”, hanya berbedanya selain mengisolasijanin juga menghancurkan janin dengan formula methotrexate dikandungdidalamnya. Formula methotrexate ini sangat ampuh untuk menghancurkan janinmenjadi serpihan-serpihan kecil akan sangat berguna pada saat dikeluarkan nanti. APA ALASAN WANITA MELAKUKAN ABORSI? Aborsi di lakukan wanita hamil baik yang sudah menikah maupun belum menikah dengan berbagai alasan , akan tetapi alasan yang utama adalah alasan-alasan non medis (termasuk aborsi sendiri / di sengaja/ buatan] MELAYANI PEMESANAN OBAT ABORSI SETIAP HARI, SIAP KIRIM KESELURUH KOTA BESAR DI INDONESIA DAN LUAR NEGERI. HUBUNGI PEMESANAN LEBIH NYAMAN VIA WA/: 085657271886

Jual Obat Aborsi Hongkong ( Asli No.1 ) 085657271886 Obat Penggugur Kandungan...

ZurliaSoop

Plant propagation: Sexual and Asexual propapagation.pptx

UmeshTimilsina1

Towards a code of practice for AI in AT.pptx

Jisc

TỔNG ÔN TẬP THI VÀO LỚP 10 MÔN TIẾNG ANH NĂM HỌC 2023 - 2024 CÓ ĐÁP ÁN (NGỮ Â...

Nguyen Thanh Tu Collection

How to Add New Custom Addons Path in Odoo 17

Celine George

NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...

Amil baba

Food safety_Challenges food safety laboratories_.pdf

Sherif Taha

Key note speaker Neum_Admir Softic_ENG.pdf

Admir Softic

COMMUNICATING NEGATIVE NEWS - APPROACHES .pptx

annathomasp01

How to Give a Domain for a Field in Odoo 17

Celine George

Recently uploaded (20)

How to setup Pycharm environment for Odoo 17.pptx

On National Teacher Day, meet the 2024-25 Kenan Fellows

How to Manage Global Discount in Odoo 17 POS

On_Translating_a_Tamil_Poem_by_A_K_Ramanujan.pptx

This PowerPoint helps students to consider the concept of infinity.

FSB Advising Checklist - Orientation 2024

General Principles of Intellectual Property: Concepts of Intellectual Proper...

Fostering Friendships - Enhancing Social Bonds in the Classroom

Sensory_Experience_and_Emotional_Resonance_in_Gabriel_Okaras_The_Piano_and_Th...

Mehran University Newsletter Vol-X, Issue-I, 2024

Jual Obat Aborsi Hongkong ( Asli No.1 ) 085657271886 Obat Penggugur Kandungan...

Plant propagation: Sexual and Asexual propapagation.pptx

Towards a code of practice for AI in AT.pptx

TỔNG ÔN TẬP THI VÀO LỚP 10 MÔN TIẾNG ANH NĂM HỌC 2023 - 2024 CÓ ĐÁP ÁN (NGỮ Â...

How to Add New Custom Addons Path in Odoo 17

NO1 Top Black Magic Specialist In Lahore Black magic In Pakistan Kala Ilam Ex...

Food safety_Challenges food safety laboratories_.pdf

Key note speaker Neum_Admir Softic_ENG.pdf

COMMUNICATING NEGATIVE NEWS - APPROACHES .pptx

How to Give a Domain for a Field in Odoo 17

#includeiostream#includestdlib.husing namespace std;class .pdf

1. #include #include using namespace std; class twoStacks { int *arr; int size; int top1, top2; public: twoStacks(int n) // constructor { size = n; arr = new int[n]; top1 = -1; top2 = size; } // Method to push an element x to stack1 void push1(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top1++; arr[top1] = x; } else { cout << "Stack Overflow"; exit(1); } } // Method to push an element x to stack2 void push2(int x) { // There is at least one empty space for new element

2. if (top1 < top2 - 1) { top2--; arr[top2] = x; } else { cout << "Stack Overflow"; exit(1); } } // Method to pop an element from first stack int pop1() { if (top1 >= 0 ) { int x = arr[top1]; top1--; return x; } else { cout << "Stack UnderFlow"; exit(1); } } // Method to pop an element from second stack int pop2() { if (top2 < size) { int x = arr[top2]; top2++; return x; } else

3. { cout << "Stack UnderFlow"; exit(1); } } }; /* Driver program to test twStacks class */ int main() { twoStacks ts(5); ts.push1(5); ts.push2(10); ts.push2(15); ts.push1(11); ts.push2(7); cout << "Popped element from stack1 is " << ts.pop1(); ts.push2(40); cout << " Popped element from stack2 is " << ts.pop2(); return 0; } #include #include using namespace std; class twoStacks { int *arr; int size; int top1, top2; public: twoStacks(int n) // constructor { size = n; arr = new int[n]; top1 = -1; top2 = size; }

4. // Method to push an element x to stack1 void push1(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top1++; arr[top1] = x; } else { cout << "Stack Overflow"; exit(1); } } // Method to push an element x to stack2 void push2(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top2--; arr[top2] = x; } else { cout << "Stack Overflow"; exit(1); } } // Method to pop an element from first stack int pop1() { if (top1 >= 0 ) { int x = arr[top1];

5. top1--; return x; } else { cout << "Stack UnderFlow"; exit(1); } } // Method to pop an element from second stack int pop2() { if (top2 < size) { int x = arr[top2]; top2++; return x; } else { cout << "Stack UnderFlow"; exit(1); } } }; /* Driver program to test twStacks class */ int main() { twoStacks ts(5); ts.push1(5); ts.push2(10); ts.push2(15); ts.push1(11); ts.push2(7); cout << "Popped element from stack1 is " << ts.pop1(); ts.push2(40);

6. cout << " Popped element from stack2 is " << ts.pop2(); return 0; } Solution #include #include using namespace std; class twoStacks { int *arr; int size; int top1, top2; public: twoStacks(int n) // constructor { size = n; arr = new int[n]; top1 = -1; top2 = size; } // Method to push an element x to stack1 void push1(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top1++; arr[top1] = x; } else { cout << "Stack Overflow"; exit(1); }

7. } // Method to push an element x to stack2 void push2(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top2--; arr[top2] = x; } else { cout << "Stack Overflow"; exit(1); } } // Method to pop an element from first stack int pop1() { if (top1 >= 0 ) { int x = arr[top1]; top1--; return x; } else { cout << "Stack UnderFlow"; exit(1); } } // Method to pop an element from second stack int pop2() { if (top2 < size) {

8. int x = arr[top2]; top2++; return x; } else { cout << "Stack UnderFlow"; exit(1); } } }; /* Driver program to test twStacks class */ int main() { twoStacks ts(5); ts.push1(5); ts.push2(10); ts.push2(15); ts.push1(11); ts.push2(7); cout << "Popped element from stack1 is " << ts.pop1(); ts.push2(40); cout << " Popped element from stack2 is " << ts.pop2(); return 0; } #include #include using namespace std; class twoStacks { int *arr; int size; int top1, top2; public: twoStacks(int n) // constructor {

9. size = n; arr = new int[n]; top1 = -1; top2 = size; } // Method to push an element x to stack1 void push1(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top1++; arr[top1] = x; } else { cout << "Stack Overflow"; exit(1); } } // Method to push an element x to stack2 void push2(int x) { // There is at least one empty space for new element if (top1 < top2 - 1) { top2--; arr[top2] = x; } else { cout << "Stack Overflow"; exit(1); } } // Method to pop an element from first stack

10. int pop1() { if (top1 >= 0 ) { int x = arr[top1]; top1--; return x; } else { cout << "Stack UnderFlow"; exit(1); } } // Method to pop an element from second stack int pop2() { if (top2 < size) { int x = arr[top2]; top2++; return x; } else { cout << "Stack UnderFlow"; exit(1); } } }; /* Driver program to test twStacks class */ int main() { twoStacks ts(5); ts.push1(5); ts.push2(10);

11. ts.push2(15); ts.push1(11); ts.push2(7); cout << "Popped element from stack1 is " << ts.pop1(); ts.push2(40); cout << " Popped element from stack2 is " << ts.pop2(); return 0; }

#includeiostream#includestdlib.husing namespace std;class .pdf

Recommended

Recommended

More Related Content

Similar to #includeiostream#includestdlib.husing namespace std;class .pdf

Similar to #includeiostream#includestdlib.husing namespace std;class .pdf (20)

More from asif1401

More from asif1401 (20)

Recently uploaded

Recently uploaded (20)

#includeiostream#includestdlib.husing namespace std;class .pdf