Sum of all integers between two integers

Here given code implementation process.

/*
    C program for
    Sum of all integers between two integers 
*/
#include <stdio.h>

void sumBetweenAtoB(int min, int max)
{
    if (max < min)
    {
        return;
    }
    /*
        Formula
        resutl = ((n+1) *min) + ((n *(n+1))/2))
        // Here
        min : starting number
        max : last number
        n = max - min
        n indicate number of elements between min to max
        ((n *(n+1))/2) sum of natural number from 1 to n
    */
    int n = max - min;
    // Calculated sum
    int result = ((n + 1) *min) + ((n *(n + 1)) / 2);
    // Display calculated result
    printf("\n Sum of number from (%d..%d) is : %d", min, max, result);
}
int main(int argc, char
    const *argv[])
{
    // Test A
    /*
        min = 4
        max = 10
        [4+5+6+7+8+9+10] = 49
    */
    sumBetweenAtoB(4, 10);
    // Test B
    /*
        min = -5
        max = 6
        [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
    */
    sumBetweenAtoB(-5, 6);
    // Test C
    /*
        min = 55
        max = 70
        [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
    */
    sumBetweenAtoB(55, 70);
    return 0;
}

Output

 Sum of number from (4..10) is : 49
 Sum of number from (-5..6) is : 6
 Sum of number from (55..70) is : 1000
/*
    Java program for
    Sum of all integers between two integers 
*/
public class Addition
{
    public void sumBetweenAtoB(int min, int max)
    {
        if (max < min)
        {
            return;
        }
        /*
            Formula
            resutl = ((n+1) *min) + ((n *(n+1))/2))
            // Here
            min : starting number
            max : last number
            n = max - min
            n indicate number of elements between min to max
            ((n *(n+1))/2) sum of natural number from 1 to n
        */
        int n = max - min;
        // Calculated sum
        int result = ((n + 1) * min) + ((n * (n + 1)) / 2);
        // Display calculated result
        System.out.print("\n Sum of number from (" + 
                         min + ".." + max + ") is : " + result );
    }
    public static void main(String[] args)
    {
        Addition task = new Addition();

        // Test A
        /*
            min = 4
            max = 10
            [4+5+6+7+8+9+10] = 49
        */
        task.sumBetweenAtoB(4, 10);
        // Test B
        /*
            min = -5
            max = 6
            [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
        */
        task.sumBetweenAtoB(-5, 6);
        // Test C
        /*
            min = 55
            max = 70
            [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
        */
        task.sumBetweenAtoB(55, 70);

    }   
}

Output

 Sum of number from (4..10) is : 49
 Sum of number from (-5..6) is : 6
 Sum of number from (55..70) is : 1000
// Include header file
#include <iostream>

using namespace std;
/*
    C++ program for
    Sum of all integers between two integers 
*/
class Addition
{
	public: void sumBetweenAtoB(int min, int max)
	{
		if (max < min)
		{
			return;
		}
		/*
		    Formula
		    resutl = ((n+1) *min) + ((n *(n+1))/2))
		    // Here
		    min : starting number
		    max : last number
		    n = max - min
		    n indicate number of elements between min to max
		    ((n *(n+1))/2) sum of natural number from 1 to n
		*/
		int n = max - min;
		// Calculated sum
		int result = ((n + 1) *min) + ((n *(n + 1)) / 2);
		// Display calculated result
		cout << "\n Sum of number from (" << min 
             << ".." << max << ") is : " << result;
	}
};
int main()
{
	Addition *task = new Addition();
	// Test A
	/*
	    min = 4
	    max = 10
	    [4+5+6+7+8+9+10] = 49
	*/
	task->sumBetweenAtoB(4, 10);
	// Test B
	/*
	    min = -5
	    max = 6
	    [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
	*/
	task->sumBetweenAtoB(-5, 6);
	// Test C
	/*
	    min = 55
	    max = 70
	    [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
	*/
	task->sumBetweenAtoB(55, 70);
	return 0;
}

Output

 Sum of number from (4..10) is : 49
 Sum of number from (-5..6) is : 6
 Sum of number from (55..70) is : 1000
// Include namespace system
using System;
/*
    Csharp program for
    Sum of all integers between two integers 
*/
public class Addition
{
	public void sumBetweenAtoB(int min, int max)
	{
		if (max < min)
		{
			return;
		}
		/*
		    Formula
		    resutl = ((n+1) *min) + ((n *(n+1))/2))
		    // Here
		    min : starting number
		    max : last number
		    n = max - min
		    n indicate number of elements between min to max
		    ((n *(n+1))/2) sum of natural number from 1 to n
		*/
		int n = max - min;
		// Calculated sum
		int result = ((n + 1) * min) + ((n * (n + 1)) / 2);
		// Display calculated result
		Console.Write("\n Sum of number from (" + 
                      min + ".." + max + ") is : " + result);
	}
	public static void Main(String[] args)
	{
		Addition task = new Addition();
		// Test A
		/*
		    min = 4
		    max = 10
		    [4+5+6+7+8+9+10] = 49
		*/
		task.sumBetweenAtoB(4, 10);
		// Test B
		/*
		    min = -5
		    max = 6
		    [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
		*/
		task.sumBetweenAtoB(-5, 6);
		// Test C
		/*
		    min = 55
		    max = 70
		    [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
		*/
		task.sumBetweenAtoB(55, 70);
	}
}

Output

 Sum of number from (4..10) is : 49
 Sum of number from (-5..6) is : 6
 Sum of number from (55..70) is : 1000
package main
import "fmt"
/*
    Go program for
    Sum of all integers between two integers 
*/
func sumBetweenAtoB(min, max int) {
	if max < min {
		return
	}
	/*
	    Formula
	    resutl = ((n+1) *min) + ((n *(n+1))/2))
	    // Here
	    min : starting number
	    max : last number
	    n = max - min
	    n indicate number of elements between min to max
	    ((n *(n+1))/2) sum of natural number from 1 to n
	*/
	var n int = max - min
	// Calculated sum
	var result int = ((n + 1) * min) + ((n * (n + 1)) / 2)
	// Display calculated result
	fmt.Print("\n Sum of number from (", min, "..", max, ") is : ", result)
}
func main() {

	// Test A
	/*
	    min = 4
	    max = 10
	    [4+5+6+7+8+9+10] = 49
	*/
	sumBetweenAtoB(4, 10)
	// Test B
	/*
	    min = -5
	    max = 6
	    [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
	*/
	sumBetweenAtoB(-5, 6)
	// Test C
	/*
	    min = 55
	    max = 70
	    [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
	*/
	sumBetweenAtoB(55, 70)
}

Output

 Sum of number from (4..10) is : 49
 Sum of number from (-5..6) is : 6
 Sum of number from (55..70) is : 1000
// Rust program for
// Sum of all integers between two integers 
fn main(){
	// Test A
	/*
	    min = 4
	    max = 10
	    [4+5+6+7+8+9+10] = 49
	*/
	sum_between_ato_b(4, 10);
	// Test B
	/*
	    min = -5
	    max = 6
	    [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
	*/
	sum_between_ato_b(-5, 6);
	// Test C
	/*
	    min = 55
	    max = 70
	    [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
	*/
	sum_between_ato_b(55, 70);
}
fn sum_between_ato_b(min: i32, max: i32) {
	if max < min {
		return;
	}
	/*
	    Formula
	    resutl = ((n+1) *min) + ((n *(n+1))/2))
	    // Here
	    min : starting number
	    max : last number
	    n = max - min
	    n indicate number of elements between min to max
	    ((n *(n+1))/2) sum of natural number from 1 to n
	*/
	let n: i32 = max - min;
	// Calculated sum
	let result: i32 = ((n + 1) * min) + ((n * (n + 1)) / 2);
	// Display calculated result
	print!("\n Sum of number from ({}..{}) is : {}", min, max, result);
}

Output

 Sum of number from (4..10) is : 49
 Sum of number from (-5..6) is : 6
 Sum of number from (55..70) is : 1000
<?php
/*
    Php program for
    Sum of all integers between two integers 
*/
class Addition
{
	public	function sumBetweenAtoB($min, $max)
	{
		if ($max < $min)
		{
			return;
		}
		/*
		    Formula
		    resutl = ((n+1) *min) + ((n *(n+1))/2))
		    // Here
		    min : starting number
		    max : last number
		    n = max - min
		    n indicate number of elements between min to max
		    ((n *(n+1))/2) sum of natural number from 1 to n
		*/
		$n = $max - $min;
		// Calculated sum
		$result = (($n + 1) * $min) + ((int)(($n * ($n + 1)) / 2));
		// Display calculated result
		echo("\n Sum of number from (".$min.
			"..".$max.
			") is : ".$result);
	}
}

function main()
{
	$task = new Addition();
	// Test A
	/*
	    min = 4
	    max = 10
	    [4+5+6+7+8+9+10] = 49
	*/
	$task->sumBetweenAtoB(4, 10);
	// Test B
	/*
	    min = -5
	    max = 6
	    [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
	*/
	$task->sumBetweenAtoB(-5, 6);
	// Test C
	/*
	    min = 55
	    max = 70
	    [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
	*/
	$task->sumBetweenAtoB(55, 70);
}
main();

Output

 Sum of number from (4..10) is : 49
 Sum of number from (-5..6) is : 6
 Sum of number from (55..70) is : 1000
/*
    Node JS program for
    Sum of all integers between two integers 
*/
class Addition
{
	sumBetweenAtoB(min, max)
	{
		if (max < min)
		{
			return;
		}
		/*
		    Formula
		    resutl = ((n+1) *min) + ((n *(n+1))/2))
		    // Here
		    min : starting number
		    max : last number
		    n = max - min
		    n indicate number of elements between min to max
		    ((n *(n+1))/2) sum of natural number from 1 to n
		*/
		var n = max - min;
		// Calculated sum
		var result = ((n + 1) * min) + (parseInt((n * (n + 1)) / 2));
		// Display calculated result
		process.stdout.write("\n Sum of number from (" + 
                             min + ".." + max + ") is : " + result);
	}
}

function main()
{
	var task = new Addition();
	// Test A
	/*
	    min = 4
	    max = 10
	    [4+5+6+7+8+9+10] = 49
	*/
	task.sumBetweenAtoB(4, 10);
	// Test B
	/*
	    min = -5
	    max = 6
	    [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
	*/
	task.sumBetweenAtoB(-5, 6);
	// Test C
	/*
	    min = 55
	    max = 70
	    [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
	*/
	task.sumBetweenAtoB(55, 70);
}
main();

Output

 Sum of number from (4..10) is : 49
 Sum of number from (-5..6) is : 6
 Sum of number from (55..70) is : 1000
#    Python 3 program for
#    Sum of all integers between two integers 
class Addition :
	def sumBetweenAtoB(self, min, max) :
		if (max < min) :
			return
		
		#    Formula
		#    resutl = ((n+1) *min) + ((n *(n+1))/2))
		#    // Here
		#    min : starting number
		#    max : last number
		#    n = max - min
		#    n indicate number of elements between min to max
		#    ((n *(n+1))/2) sum of natural number from 1 to n
		n = max - min
		#  Calculated sum
		result = ((n + 1) * min) + (int((n * (n + 1)) / 2))
		#  Display calculated result
		print("\n Sum of number from (", 
              min ,"..", max ,") is : ", result, end = "")
	

def main() :
	task = Addition()
	#  Test A
	#    min = 4
	#    max = 10
	#    [4+5+6+7+8+9+10] = 49
	task.sumBetweenAtoB(4, 10)
	#  Test B
	#    min = -5
	#    max = 6
	#    [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
	task.sumBetweenAtoB(-5, 6)
	#  Test C
	#    min = 55
	#    max = 70
	#    [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
	task.sumBetweenAtoB(55, 70)

if __name__ == "__main__": main()

Output

 Sum of number from ( 4 .. 10 ) is :  49
 Sum of number from ( -5 .. 6 ) is :  6
 Sum of number from ( 55 .. 70 ) is :  1000
#    Ruby program for
#    Sum of all integers between two integers 
class Addition 
	def sumBetweenAtoB(min, max) 
		if (max < min) 
			return
		end

		#    Formula
		#    resutl = ((n+1) *min) + ((n *(n+1))/2))
		#    // Here
		#    min : starting number
		#    max : last number
		#    n = max - min
		#    n indicate number of elements between min to max
		#    ((n *(n+1))/2) sum of natural number from 1 to n
		n = max - min
		#  Calculated sum
		result = ((n + 1) * min) + ((n * (n + 1)) / 2)
		#  Display calculated result
		print("\n Sum of number from (", 
              min ,"..", max ,") is : ", result)
	end

end

def main() 
	task = Addition.new()
	#  Test A
	#    min = 4
	#    max = 10
	#    [4+5+6+7+8+9+10] = 49
	task.sumBetweenAtoB(4, 10)
	#  Test B
	#    min = -5
	#    max = 6
	#    [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
	task.sumBetweenAtoB(-5, 6)
	#  Test C
	#    min = 55
	#    max = 70
	#    [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
	task.sumBetweenAtoB(55, 70)
end

main()

Output

 Sum of number from (4..10) is : 49
 Sum of number from (-5..6) is : 6
 Sum of number from (55..70) is : 1000
/*
    Scala program for
    Sum of all integers between two integers 
*/
class Addition()
{
	def sumBetweenAtoB(min: Int, max: Int): Unit = {
		if (max < min)
		{
			return;
		}
		/*
		    Formula
		    resutl = ((n+1) *min) + ((n *(n+1))/2))
		    // Here
		    min : starting number
		    max : last number
		    n = max - min
		    n indicate number of elements between min to max
		    ((n *(n+1))/2) sum of natural number from 1 to n
		*/
		var n: Int = max - min;
		// Calculated sum
		var result: Int = ((n + 1) * min) + ((n * (n + 1)) / 2);
		// Display calculated result
		print("\n Sum of number from (" + 
              min + ".." + max + ") is : " + result);
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Addition = new Addition();
		// Test A
		/*
		    min = 4
		    max = 10
		    [4+5+6+7+8+9+10] = 49
		*/
		task.sumBetweenAtoB(4, 10);
		// Test B
		/*
		    min = -5
		    max = 6
		    [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
		*/
		task.sumBetweenAtoB(-5, 6);
		// Test C
		/*
		    min = 55
		    max = 70
		    [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
		*/
		task.sumBetweenAtoB(55, 70);
	}
}

Output

 Sum of number from (4..10) is : 49
 Sum of number from (-5..6) is : 6
 Sum of number from (55..70) is : 1000
/*
    Swift 4 program for
    Sum of all integers between two integers 
*/
class Addition
{
	func sumBetweenAtoB(_ min: Int, _ max: Int)
	{
		if (max < min)
		{
			return;
		}
		/*
		    Formula
		    resutl = ((n+1) *min) + ((n *(n+1))/2))
		    // Here
		    min : starting number
		    max : last number
		    n = max - min
		    n indicate number of elements between min to max
		    ((n *(n+1))/2) sum of natural number from 1 to n
		*/
		let n: Int = max - min;
		// Calculated sum
		let result: Int = ((n + 1) * min) + ((n * (n + 1)) / 2);
		// Display calculated result
		print("\n Sum of number from (", 
              min ,"..", max ,") is : ", 
              result, terminator: "");
	}
}
func main()
{
	let task: Addition = Addition();
	// Test A
	/*
	    min = 4
	    max = 10
	    [4+5+6+7+8+9+10] = 49
	*/
	task.sumBetweenAtoB(4, 10);
	// Test B
	/*
	    min = -5
	    max = 6
	    [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
	*/
	task.sumBetweenAtoB(-5, 6);
	// Test C
	/*
	    min = 55
	    max = 70
	    [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
	*/
	task.sumBetweenAtoB(55, 70);
}
main();

Output

 Sum of number from ( 4 .. 10 ) is :  49
 Sum of number from ( -5 .. 6 ) is :  6
 Sum of number from ( 55 .. 70 ) is :  1000
/*
    Kotlin program for
    Sum of all integers between two integers 
*/
class Addition
{
	fun sumBetweenAtoB(min: Int, max: Int): Unit
	{
		if (max < min)
		{
			return;
		}
		/*
		    Formula
		    resutl = ((n+1) *min) + ((n *(n+1))/2))
		    // Here
		    min : starting number
		    max : last number
		    n = max - min
		    n indicate number of elements between min to max
		    ((n *(n+1))/2) sum of natural number from 1 to n
		*/
		val n: Int = max - min;
		// Calculated sum
		val result: Int = ((n + 1) * min) + ((n * (n + 1)) / 2);
		// Display calculated result
		print("\n Sum of number from (" + 
              min + ".." + max + ") is : " + result);
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Addition = Addition();
	// Test A
	/*
	    min = 4
	    max = 10
	    [4+5+6+7+8+9+10] = 49
	*/
	task.sumBetweenAtoB(4, 10);
	// Test B
	/*
	    min = -5
	    max = 6
	    [-5+-4+-3+-2+-1+0+1+2+3+4+5+6] = 6
	*/
	task.sumBetweenAtoB(-5, 6);
	// Test C
	/*
	    min = 55
	    max = 70
	    [55+56+57+58+59+60+61+62+63+64+65+66+67+68+69+70] = 1000
	*/
	task.sumBetweenAtoB(55, 70);
}

Output

 Sum of number from (4..10) is : 49
 Sum of number from (-5..6) is : 6
 Sum of number from (55..70) is : 1000


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