Check whether a number can be represented as sum of K distinct positive integers

Given two integers N and K, the task is to check whether N can be represented as sum of K distinct positive integers.

Examples:

Input: N = 12, K = 4
Output: Yes
N = 1 + 2 + 4 + 5 = 12 (12 as sum of 4 distinct integers)



Input: N = 8, K = 4
Output: No

Approach: Consider the series 1 + 2 + 3 + … + K which has exactly K distinct integers with minimum possible sum i.e. Sum = (K * (K – 1)) / 2. Now, if N < Sum then it is not possible to represent N as the sum of K distinct positive integers but if N ≥ Sum then any integer say X ≥ 0 can be added to Sum to generate the sum equal to N i.e. 1 + 2 + 3 + … + (K – 1) + (K + X) ensuring that there are exactly K distinct positive integers.

Below is the implementation of the above approach:

C++

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// 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;
}

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Java














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// 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.. 



















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Python3














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# 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 



















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C#














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// 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. 



















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PHP














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<?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 


  


?> 



















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Output:


Yes





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