sorting

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Wed Dec 09 2020 03:00:16 GMT+0000 (UTC)

Saved by @mahmoud hussein #c++

 
#include<iostream>
#include <fstream>
using namespace std;
void selectionSort(double list[], int listSize);
int main()
{
    double list[] = { 3.4, 5, 3, 3.5, 2.2, 1.9, 2 };

    selectionSort(list, 7);
    for (int i = 0; i < 7; i++)
    {
        cout << list[i]<<" ";
    }
}
//decreasing order
void selectionSort(double list[], int listSize)
{
    for (int i = listSize-1; i >= 0 ; i--)
    {
        double currentMax = list[i];
        int currentMaxIndex = i;
        for (int j = i - 1; j >= 0; j--)
        {
            if (currentMax < list[j])
            {
                currentMax = list[j];
                currentMaxIndex = j;
            }
        }
            if (currentMaxIndex != i)
            {
                list[currentMaxIndex] = list[i];
                list[i] = currentMax;
            }
   
    }
}
//increasing order
void selectionSort(double list[], int listSize)
{
    for (int i = 0; i < listSize - 1; i++)
    {
        double currentMin = list[i];
        int currentMinIndex = i;
        for (int j = i + 1; j < listSize; j++)
        {
            if (currentMin > list[j])
            {
                currentMin = list[j];
                currentMinIndex = j;
            }
        }
            if (currentMinIndex != i)
            {
                list[currentMinIndex] = list[i];
                list[i] = currentMin;
            }
        
    }
}

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c++ loops

@mahmoud hussein

(Occurrence of max numbers) Write a program that reads integers, finds the largest of them, and counts its occurrences. Assume that the input ends with number 0. Suppose that you entered 3 5 2 5 5 5 0; the program finds that the largest is 5 and the occurrence count for 5 is 4 (Fibonacci Series) The Fibonacci series is a series that begins with 0 and 1 and has the property that each succeeding term is the sum of the two preceding terms. For example, the third Fibonacci number is 1 which is sum of 0 and 1. The next is 2, which is a sum of 1 + 1. Write a program that displays the first ten numbers in a Fibonacci series **5.21 (Display numbers in a pyramid pattern) Write a nested for loop that prints the following output: (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. sorting

#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;    
 }    
   return 0;  
}

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 );
    return 0;
}

(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;

	
	
		

	
	
}

  
#include <ESP8266WiFi.h>


// Set these to run example.
#define WIFI_SSID "SSID"
#define WIFI_PASSWORD "PASSWORD"

  // connect to wifi.
  WiFi.begin(WIFI_SSID, WIFI_PASSWORD);
  Serial.print("connecting");
  while (WiFi.status() != WL_CONNECTED) {
    Serial.print(".");
    delay(500);
  }
  Serial.println();
  Serial.print("connected: ");
  Serial.println(WiFi.localIP());

(Display the ASCII character table) Write a program that prints all the uppercase characters of the ASCII character table. Display 5 characters per line. The ASCII table is shown in Appendix B. Characters are separated by exactly one space.

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

using namespace std;

int main()
{
	int firstLetter = 65;
	int count = 0;

	for (int i = 65; i <= 90; i++)
	{
		count++;
		if (count%5 == 0)
			cout << setw(3) << static_cast<char>(i) << endl;
		else
			cout << setw(3) << static_cast<char>(i);

		
		
	}

	
	
		

	
	
}

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

		}
		for (int j = 7; j >= i; j--)
		{
			cout << " " << static_cast<char>(j + 97 - i);
		}
			cout << "\n";
	}
		

}

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++)
        {
            cout << v<<"  ";
        }
        cout << endl;

        
    }


}

firebase basics

include <FirebaseArduino.h>

#define FIREBASE_HOST "example.firebaseio.com"
#define FIREBASE_AUTH "token_or_secret"


void setup() {
  Serial.begin(9600);
  Firebase.begin(FIREBASE_HOST, FIREBASE_AUTH);
}

int n = 0;

void loop() {

  // set value
  Firebase.setFloat("number", 42.0);
  // handle error
  if (Firebase.failed()) {
      Serial.print("setting /number failed:");
      Serial.println(Firebase.error());  
      return;
  }
  delay(1000);
  
  // update value
  Firebase.setFloat("number", 43.0);
  // handle error
  if (Firebase.failed()) {
      Serial.print("setting /number failed:");
      Serial.println(Firebase.error());  
      return;
  }
  delay(1000);

  // get value 
  Serial.print("number: ");
  Serial.println(Firebase.getFloat("number"));
  delay(1000);

  // remove value
  Firebase.remove("number");
  delay(1000);

  // set string value
  Firebase.setString("message", "hello world");
  // handle error
  if (Firebase.failed()) {
      Serial.print("setting /message failed:");
      Serial.println(Firebase.error());  
      return;
  }
  delay(1000);
  
  // set bool value
  Firebase.setBool("truth", false);
  // handle error
  if (Firebase.failed()) {
      Serial.print("setting /truth failed:");
      Serial.println(Firebase.error());  
      return;
  }
  delay(1000);

  // append a new value to /logs
  String name = Firebase.pushInt("logs", n++);
  // handle error
  if (Firebase.failed()) {
      Serial.print("pushing /logs failed:");
      Serial.println(Firebase.error());  
      return;
  }
  Serial.print("pushed: /logs/");
  Serial.println(name);
  delay(1000);
}

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