Find sum of modulo K of first N natural number

Given two integer N ans K, the task is to find sum of modulo K of first N natural numbers i.e 1%K + 2%K + ….. + N%K.

Examples :

Input : N = 10 and K = 2.
Output : 5
Sum = 1%2 + 2%2 + 3%2 + 4%2 + 5%2 + 6%2 +
      7%2 + 8%2 + 9%2 + 10%2
   = 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0
   = 5.

Method 1:
Iterate a varible i from 1 to N, evaluate and add i%K.

Below is the implementation of this approach:

C++

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// C++ program to find sum of
// modulo K of first N natural numbers.
#include <bits/stdc++.h>
using namespace std;
  
// Return sum of modulo K of
// first N natural numbers.
int findSum(int N, int K)
{
    int ans = 0;
  
    // Iterate from 1 to N &&
    // evaluating and adding i % K.
    for (int i = 1; i <= N; i++)
        ans += (i % K);
  
    return ans;
}
  
// Driver Program
int main()
{
    int N = 10, K = 2;
    cout << findSum(N, K) << endl;
    return 0;
}

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Java

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// Java program to find sum of modulo
// K of first N natural numbers.
import java.io.*;
  
class GFG {
  
    // Return sum of modulo K of
    // first N natural numbers.
    static int findSum(int N, int K)
    {
        int ans = 0;
  
        // Iterate from 1 to N && evaluating
        // and adding i % K.
        for (int i = 1; i <= N; i++)
            ans += (i % K);
  
        return ans;
    }
  
    // Driver program
    static public void main(String[] args)
    {
        int N = 10, K = 2;
        System.out.println(findSum(N, K));
    }
}
  
// This code is contributed by vt_m.

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Python3

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# Python3 program to find sum 
# of modulo K of first N 
# natural numbers.
  
# Return sum of modulo K of 
# first N natural numbers.
  
def findSum(N, K):
    ans = 0;
  
    # Iterate from 1 to N &&
    # evaluating and adding i % K.
    for i in range(1, N + 1):
        ans += (i % K);
  
    return ans;
  
# Driver Code
N = 10
K = 2;
print(findSum(N, K));
  
# This code is contributed by mits

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C#

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// C# program to find sum of modulo
// K of first N natural numbers.
using System;
  
class GFG {
  
    // Return sum of modulo K of
    // first N natural numbers.
    static int findSum(int N, int K)
    {
        int ans = 0;
  
        // Iterate from 1 to N && evaluating
        // and adding i % K.
        for (int i = 1; i <= N; i++)
            ans += (i % K);
  
        return ans;
    }
  
    // Driver program
    static public void Main()
    {
        int N = 10, K = 2;
        Console.WriteLine(findSum(N, K));
    }
}
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP program to find sum 
// of modulo K of first N 
// natural numbers.
  
  
// Return sum of modulo K of 
// first N natural numbers.
  
function findSum($N, $K)
{
    $ans = 0;
  
    // Iterate from 1 to N &&
    // evaluating and adding i % K.
    for ($i = 1; $i <= $N; $i++)
        $ans += ($i % $K);
  
    return $ans;
}
  
// Driver Code
$N = 10; $K = 2;
echo findSum($N, $K), "\n";
  
// This code is contributed by ajit
?>

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Output :

5

Time Complexity : O(N).

Method 2 :
Two cases arise in this method.

Case 1: When N < K, for each number i, N >= i >= 1, will give i as result when operate with modulo K. So, required sum will be sum of first N natural number, N*(N+1)/2.

Case 2: When N >= K, then integers from 1 to K in natural number sequence will produce, 1, 2, 3, ….., K – 1, 0 as result when operate with modulo K. Similarly, from K + 1 to 2K, it will produce same result. So, the idea is to count how many number of times this sequence appears and multiply it with sum of first K – 1 natural numbers.
Below is the implementation of this approach:

C++

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// C++ program to find sum of modulo
// K of first N natural numbers.
#include <bits/stdc++.h>
using namespace std;
  
// Return sum of modulo K of
// first N natural numbers.
int findSum(int N, int K)
{
    int ans = 0;
  
    // Counting the number of times 1, 2, ..,
    // K-1, 0 sequence occurs.
    int y = N / K;
  
    // Finding the number of elements left which
    // are incomplete of sequence Leads to Case 1 type.
    int x = N % K;
  
    // adding multiplication of number of
    // times 1, 2, .., K-1, 0 sequence occurs
    // and sum of first k natural number and sequence
    // from case 1.
    ans = (K * (K - 1) / 2) * y + (x * (x + 1)) / 2;
  
    return ans;
}
  
// Driver program
int main()
{
    int N = 10, K = 2;
    cout << findSum(N, K) << endl;
    return 0;
}

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Java

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// Java program to find sum of modulo
// K of first N natural numbers.
import java.io.*;
  
class GFG {
  
    // Return sum of modulo K of
    // first N natural numbers.
    static int findSum(int N, int K)
    {
        int ans = 0;
  
        // Counting the number of times 1, 2, ..,
        // K-1, 0 sequence occurs.
        int y = N / K;
  
        // Finding the number of elements left which
        // are incomplete of sequence Leads to Case 1 type.
        int x = N % K;
  
        // adding multiplication of number of times
        // 1, 2, .., K-1, 0 sequence occurs and sum
        // of first k natural number and sequence
        // from case 1.
        ans = (K * (K - 1) / 2) * y + (x * (x + 1)) / 2;
  
        return ans;
    }
  
    // Driver program
    static public void main(String[] args)
    {
        int N = 10, K = 2;
        System.out.println(findSum(N, K));
    }
}
  
// This Code is contributed by vt_m.

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Python3

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# Python3 program to find sum of modulo
# K of first N natural numbers.
  
# Return sum of modulo K of
# first N natural numbers.
def findSum(N, K):
  
    ans = 0;
  
    # Counting the number of times
    # 1, 2, .., K-1, 0 sequence occurs.
    y = N / K;
  
    # Finding the number of elements
    # left which are incomplete of 
    # sequence Leads to Case 1 type.
    x = N % K;
  
    # adding multiplication of number
    # of times 1, 2, .., K-1, 0 
    # sequence occurs and sum of 
    # first k natural number and 
    # sequence from case 1.
    ans = ((K * (K - 1) / 2) * y + 
                (x * (x + 1)) / 2);
  
    return int(ans);
  
# Driver Code
N = 10;
K = 2;
print(findSum(N, K));
  
# This code is contributed by mits

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C#

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// C# program to find sum of modulo
// K of first N natural numbers.
using System;
  
class GFG {
  
    // Return sum of modulo K of
    // first N natural numbers.
    static int findSum(int N, int K)
    {
        int ans = 0;
  
        // Counting the number of times 1, 2, ..,
        // K-1, 0 sequence occurs.
        int y = N / K;
  
        // Finding the number of elements left which
        // are incomplete of sequence Leads to Case 1 type.
        int x = N % K;
  
        // adding multiplication of number of times
        // 1, 2, .., K-1, 0 sequence occurs and sum
        // of first k natural number and sequence
        // from case 1.
        ans = (K * (K - 1) / 2) * y + (x * (x + 1)) / 2;
  
        return ans;
    }
  
    // Driver program
    static public void Main()
    {
        int N = 10, K = 2;
        Console.WriteLine(findSum(N, K));
    }
}
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP program to find sum of modulo
// K of first N natural numbers.
  
// Return sum of modulo K of
// first N natural numbers.
function findSum($N, $K)
{
    $ans = 0;
  
    // Counting the number of times
    // 1, 2, .., K-1, 0 sequence occurs.
    $y = $N / $K;
  
    // Finding the number of elements
    // left which are incomplete of 
    // sequence Leads to Case 1 type.
    $x = $N % $K;
  
    // adding multiplication of number
    // of times 1, 2, .., K-1, 0 
    // sequence occurs and sum of 
    // first k natural number and 
    // sequence from case 1.
    $ans = ($K * ($K - 1) / 2) * $y
                 + ($x * ($x + 1)) / 2;
  
    return $ans;
}
  
// Driver program
    $N = 10; $K = 2;
    echo findSum($N, $K) ;
  
// This code is contributed by anuj_67.
?>

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Output :

5

Time Complexity : O(1).

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : vt_m, jit_t, Mithun Kumar