(Count occurrence of numbers) Write a program that reads at most 100 integers between 1 and 100 and counts the occurrence of each number. Assume the input ends with 0. Here is a sample run of the program:

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Fri Dec 11 2020 21:08:00 GMT+0000 (UTC)

Saved by @mahmoud hussein #c++

 int list[LIMIT], i, j, val, count, flag, test[LIMIT], c = 0;
    //enter the values
    for (i = 0; i < LIMIT; i++)
    {
        cout << "Enter The Elements : ";
        cin >> list[i];
        if (list[i] == 0)
            break;
    }
    //count the occurence of each value
    for (j = 0; j < LIMIT; j++)
    {
        count = 0;
        flag = 1;
        val = list[j];
        //counts each value's occurence and if the value is 0 break
        for (i = 0; i < LIMIT; i++)
        {
            if (val == list[i])
                count++;
            else if (list[i] == 0)
                break;
        }
        //checks whether the value has been already counted (flag = 0) or not (flag stays 1)
        for (i = 0; i < c; i++)
        {
            if (test[i] == val)
                flag = 0;
        }
        //if the value has not been counted print 
        if (flag == 1)
        {

            cout << "The Occurence Of The Number " << val << " is " << count << endl;
            test[c] = val;
            c++;
        }

    }

http://cpp.sh/

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c++ arrays

@mahmoud hussein

pulling 4 random cards from a deck (Count the number of Vowels and Consonants) (Count occurrence of numbers) Write a program that reads at most 100 integers between 1 and 100 and counts the occurrence of each number. Assume the input ends with 0. Here is a sample run of the program: (Game: Eight Queens) 8 (Partition of a list) CheckSudokuSolution sorting td arrays (Financial application: compute tax) **8.11 (Game: nine heads and tails) (Pattern recognition: consecutive four equal numbers) Write the following function that tests whether a two-dimensional array has four consecutive numbers of the same value, either horizontally, vertically, or diagonally. (Game: connect four) Connect four is a two-player board game in which the players alternately drop colored disks into a seven-column, six-row verticallysuspended grid (Game: connect four) Connect four is a two-player board game in which the players alternately drop colored disks into a seven-column, six-row verticallysuspended grid

#include<iostream>
using namespace std;

int sum(int arr[], int a);

int main() {
    int arr[1000], n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> arr[i];
        cout << sum(arr, n);
    }
}
int sum(int arr[], int a) {
    int i;
    sum = 0;
    for (i=0; i<n; ++i) {
        sum += arr[i];
    }
    return sum;
}#include<iostream>
using namespace std;

int sum(int arr[], int a);

int main() {
    int arr[1000], n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        cin >> arr[i];
        cout << sum(arr, n);
    }
}
int sum(int arr[], int a) {
    int i;
    sum = 0;
    for (i=0; i<n; ++i) {
        sum += arr[i];
    }
    return sum;
}

#c++ #algorithms #interviewquestions #arrays

Kadane’s Algorithm

int maxSubArraySum(int a[], int size) 
{ 
int max_so_far = 0, max_ending_here = 0; 
for (int i = 0; i < size; i++) 
{ 
	max_ending_here = max_ending_here + a[i]; 
	if (max_ending_here < 0) 
		max_ending_here = 0; 

	/* Do not compare for all elements. Compare only 
		when max_ending_here > 0 */
	else if (max_so_far < max_ending_here) 
		max_so_far = max_ending_here; 
} 
return max_so_far; 
}

#c++ #fibonacci

Fibonacci Series(asking user for number of elements)

#include <iostream>  
using namespace std;  
int main() 
{  
  int n1=0,n2=1,n3,i,number;    
 cout<<"Enter the number of elements: ";    
 cin>>number;    
 cout<<n1<<" "<<n2<<" "; //printing 0 and 1    
 for(i=2;i<number;++i) //loop starts from 2 because 0 and 1 are already printed    
 {    
  n3=n1+n2;    
  cout<<n3<<" ";    
  n1=n2;    
  n2=n3;    
 }    
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}

A Simple Binary Tree Implementation in C++

#include <stdio.h>
 
class Node
{
public:
    Node( int v )
    {
        data = v;
        left = 0;
        right = 0;
    }
 
    int data;
    Node* left;
    Node* right;
};
 
void Add( Node** root, Node* n )
{
    if ( !*root  )
    {
        *root = n;
        return;
    }
 
    if ( (*root)->data < n->data )
    {
        Add( &(*root)->right, n );
    }
    else
    {
        Add( &(*root)->left, n );
    }
}
 
void Print( Node* node )
{
    if ( !node )  return;   
    Print( node->left );
    printf( "value = %i\n", node->data );
    Print( node->right );
}
 
int main()
{
    Node* root = 0;
 
    Add( &root, new Node( 1 ) );
    Add( &root, new Node( 2 ) );
    Add( &root, new Node( -1 ) );
    Add( &root, new Node( 12 ) );
 
    Print( root );
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}

(Find the largest n such that 2n < 30,000)) Use a while loop to find the largest integer n such that 2n is less than 30,000.

#include <iostream>
#include<cmath>
#include<ctime>
#include<string>
#include <iomanip>
#include <fstream>

using namespace std;

int main()
{
	
	int n = 1;
	int largest = 0;
	while (n*n<30000)
	{
		if (n > largest)
		{
			largest = n;
		}
		n++;
	}
	cout << "the largest number is " << largest;
	
	
		

	
	
}

#include<iostream>
using namespace std;

int main()
{
  cout<<"Enter the integers: "<
}

#c++ #web #rest

GitHub - an-tao/drogon: Drogon: A C 14/17 based HTTP web application framework running on Linux/macOS/Unix/Windows

https://github.com/an-tao/drogon

(Find numbers divisible by 3 or 6, but not both) Write a program that displays all the numbers from 300 to 400, 5 per line, that are divisible by 3 or 6, but not both. Numbers are separated by exactly one space.

#include <iostream>
#include<cmath>
#include<ctime>
#include<string>
#include <iomanip>
#include <fstream>

using namespace std;

int main()
{
	
	int count = 0;
	
	// iterate from 0 to N 
	for (int num = 300,count = 0; num <=400; num++)
	{
		// Short-circuit operator is used  
		if (num % 3 == 0 || num % 6 == 0)
		{
			cout << setw(5) << num;
			count++;
			if (count % 5 == 0)
				cout << setw(5) << num << endl;
		
		}
		
		
			
	}
		

	
	
}

Creating an variable array in C++

,

(Compute the greatest common divisor) Another solution for Listing 5.10 to find the greatest common divisor (GCD) of two integers n1 and n2 is as follows: First find d to be the minimum of n1 and n2, then check whether d, d-1, d-2, . . . , 2, or 1 is a divisor for both n1 and n2 in this order. The first such common divisor is the greatest common divisor for n1 and n2. Write a program that prompts the user to enter two positive integers and displays the GCD.

#include <iostream>
#include<cmath>
#include<ctime>
#include<string>
#include <iomanip>
#include <fstream>

using namespace std;

int main()
{

	cout << "Enter two positive numbers ";
	int n1,n2;
	cin >> n1>>n2;
	int d = 1;
	int gcd = 1;
	
	if (n1 < n2)
		d = n1;
	else
		d = n2;
	while (true)
	{
		if (n1 % d == 0 && n2 % d == 0) {
				gcd = d;
				break;
		}
			
		d--;
	}
	
	cout << "the CGD of " << n1 << " and " << n2 << " is " << gcd;

	
	
		

	
	
}

  
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#include <iostream>
#include<cmath>
#include<ctime>
#include<string>
#include <iomanip>
#include <fstream>

using namespace std;

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	for (int i = 65; i <= 90; i++)
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**5.21 (Display numbers in a pyramid pattern) Write a nested for loop that prints the following output:

#include <iostream>
#include<cmath>
#include<ctime>
#include<string>
#include <iomanip>
#include <fstream>

using namespace std;

int main()
{
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	{
		for (int k = 0; k <= i; k++)
		{
			cout << "  ";
		}
		for (int j = i; j < 8; j++)
		{
			cout <<" "<< static_cast<char>(j+97-i);

		}
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		{
			cout << " " << static_cast<char>(j + 97 - i);
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			cout << "\n";
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9 (Display pyramid) Write a program that prompts the user to enter an integer from 1 to 15 and displays a pyramid, as shown in the following sample run

#include <iostream>
#include<cmath>
#include<ctime>
#include<string>
#include <iomanip>
#include <fstream>

using namespace std;

int main()
{



    int rows;
    cin >> rows;

    for (int i = 1; i <= rows; i++)
    {
        for (int k = 1; k <= (rows -i); k++)
        {
            cout << "   ";
           
        }
        for (int d = i; d > 1; d--)
            {
            cout << d << "  ";
                
            }
        for (int  v = 1; v <= i; v++)
        {
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        cout << endl;

        
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}

firebase basics

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      Serial.println(Firebase.error());  
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      Serial.println(Firebase.error());  
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  Serial.println(Firebase.getFloat("number"));
  delay(1000);

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      Serial.println(Firebase.error());  
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  Serial.print("pushed: /logs/");
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}

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