Comprobar si un número se puede representar como la suma de K enteros positivos distintos

Dados dos números enteros N y K , la tarea es verificar si N puede representarse como la suma de K números enteros positivos distintos.

Ejemplos: 

Entrada: N = 12, K = 4 
Salida: Sí 
N = 1 + 2 + 4 + 5 = 12 (12 como suma de 4 enteros distintos)

Entrada: N = 8, K = 4 
Salida: No 
 

Enfoque: considere la serie 1 + 2 + 3 + … + K que tiene exactamente K enteros distintos con la suma mínima posible, es decir, Sum = (K * (K – 1)) / 2 . Ahora, si N < Suma , entonces no es posible representar a N como la suma de K enteros positivos distintos, pero si N ≥ Suma , cualquier entero, digamos X ≥ 0 , se puede agregar a Suma para generar la suma igual a N , es decir , 1 + 2. + 3 + … + (K – 1) + (K + X) asegurando que haya exactamente K enteros positivos distintos.

A continuación se muestra la implementación del enfoque anterior:  

// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function that returns true if n
// can be represented as the sum of
// exactly k distinct positive integers
bool solve(int n, int k)
{
    // If n can be represented as
    // 1 + 2 + 3 + ... + (k - 1) + (k + x)
    if (n >= (k * (k + 1)) / 2) {
        return true;
    }
 
    return false;
}
 
// Driver code
int main()
{
    int n = 12, k = 4;
 
    if (solve(n, k))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

// Java implementation of the approach
import java.io.*;
 
class GFG {
 
    // Function that returns true if n
    // can be represented as the sum of
    // exactly k distinct positive integers
    static boolean solve(int n, int k)
    {
        // If n can be represented as
        // 1 + 2 + 3 + ... + (k - 1) + (k + x)
        if (n >= (k * (k + 1)) / 2) {
            return true;
        }
 
        return false;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 12, k = 4;
 
        if (solve(n, k))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by anuj_67..

# Python 3 implementation of the approach
 
# Function that returns true if n
# can be represented as the sum of
# exactly k distinct positive integers
def solve(n,k):
    # If n can be represented as
    # 1 + 2 + 3 + ... + (k - 1) + (k + x)
    if (n >= (k * (k + 1)) // 2):
        return True
 
    return False
 
# Driver code
if __name__ == '__main__':
    n = 12
    k = 4
 
    if (solve(n, k)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by
# Surendra_Gangwar

// C# implementation of the approach
using System;
 
class GFG
{
    // Function that returns true if n
    // can be represented as the sum of
    // exactly k distinct positive integers
    static bool solve(int n, int k)
    {
        // If n can be represented as
        // 1 + 2 + 3 + ... + (k - 1) + (k + x)
        if (n >= (k * (k + 1)) / 2) {
            return true;
        }
 
        return false;
    }
 
    // Driver code
    static public void Main ()
    {
        int n = 12, k = 4;
 
        if (solve(n, k))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by ajit.

<?php
 
// PHP implementation of the approach
 
// Function that returns true if n
// can be represented as the sum of
// exactly k distinct positive integers
function solve($n, $k)
{
    // If n can be represented as
    // 1 + 2 + 3 + ... + (k - 1) + (k + x)
    if ($n >= ($k * ($k + 1)) / 2) {
        return true;
    }
 
    return false;
}
 
// Driver code
 
$n = 12;
$k = 4;
 
if (solve($n, $k))
    echo  "Yes";
else
    echo  "No";
 
// This code is contributed by ihritik
 
?>

<script>
 
// Javascript implementation of the approach
 
// Function that returns true if n
// can be represented as the sum of
// exactly k distinct positive integers
function solve(n, k)
{
     
    // If n can be represented as
    // 1 + 2 + 3 + ... + (k - 1) + (k + x)
    if (n >= (k * (k + 1)) / 2)
    {
        return true;
    }
    return false;
}
 
// Driver code
var n = 12, k = 4;
 
if (solve(n, k))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by todaysgaurav
 
</script>

Yes

Complejidad de tiempo: O(1)

Espacio Auxiliar: O(1)