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 variable i from 1 to N, evaluate and add i%K.
Below is the implementation of this approach:
// 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;
} |
// 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. |
# 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 |
// 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. |
<?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 ?> |
<script> // JavaScript 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)
{ let ans = 0;
// Iterate from 1 to N && evaluating
// and adding i % K.
for (let i = 1; i <= N; i++)
ans += (i % K);
return ans;
} // Driver Code let N = 10, K = 2; document.write(findSum(N, K)); // This code is contributed by code_hunt </script> |
Output :
5
Time Complexity : O(N).
Auxiliary Space: O(1)
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, the required sum will be the sum of the 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 numbers of times this sequence appears and multiply it with the sum of first K – 1 natural numbers.
Below is the implementation of this approach:
// 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;
} |
// 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. |
# 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 |
// 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. |
<?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. ?> |
<script> // Javascript 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)
{ let ans = 0;
// Counting the number of times
// 1, 2, .., K-1, 0 sequence occurs.
let y = N / K;
// Finding the number of elements
// left which are incomplete of
// sequence Leads to Case 1 type.
let 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 code let N = 10; let K = 2; document.write(findSum(N, K)); // This code is contributed by _saurabh_jaiswal </script> |
Output :
5
Time Complexity : O(1).
Auxiliary Space: O(1)