Count of pairs from 1 to a and 1 to b whose sum is divisible by N

Given three integers a, b and N. Find the total number of distinct pairs which can be formed by selecting one integer from 1 to a and other from 1 to b, such that their sum is divisible by N.
Examples: 

Input : a = 4, b = 4, N = 4
Output : 4

Input : a = 5, b = 13, N = 3
Output : 22

Basic Approach: For a pair to be divisible by N it must contain one number from range 1 to a and other from 1 to b. 
So, for this iterate over integers from 1 to a and for each integer (i), b/N numbers are there whose sum with i will be divisible by N. Also if (i%N + b%N) >= N then 1 more pair exists whose sum is divisible by N.
For example took a = 7, b = 6 and N = 4: 

Let's check for i = 3:
  • b/N = 6/4 = 1 => there is one integer from 1 to b,
    whose sum with 3 is divisible by 4 i.e.(3, 1).
  • Also i%N + b%N = 3%4 + 6%4 = 3+2 = 5 > 4, 
    means one more integer exists from 1 to b 
    whose sum with 3 is divisible by 4 i.e.(3, 5).

Below is the implementation of the above approach: 
 

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
int findCountOfPairs(int a, int b, int n)
{
    int ans = 0;
 
    // Iterate over 1 to a to find distinct pairs
    for (int i = 1; i <= a; i++) {
        // For each integer from 1 to a
        // b/n integers exists such that pair
        // sum is divisible by n
        ans += b / n;
 
        // If (i%n +b%n ) >= n one more pair is possible
        ans += (i % n + b % n) >= n ? 1 : 0;
    }
 
    // Return answer
    return ans;
}
 
// Driver code
int main()
{
    int a = 5, b = 13, n = 3;
    cout << findCountOfPairs(a, b, n);
 
    return 0;
}

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Java

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// Java program for the above approach
class GFG
{
 
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
static int findCountOfPairs(int a, int b, int n)
{
    int ans = 0;
 
    // Iterate over 1 to a to find distinct pairs
    for (int i = 1; i <= a; i++)
    {
        // For each integer from 1 to a
        // b/n integers exists such that pair
        // sum is divisible by n
        ans += b / n;
 
        // If (i%n +b%n ) >= n one more pair is possible
        ans += (i % n + b % n) >= n ? 1 : 0;
    }
 
    // Return answer
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int a = 5, b = 13, n = 3;
    System.out.println(findCountOfPairs(a, b, n));
}
}
 
// This code has been contributed by 29AjayKumar

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Python3

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# Python implementation of above approach
 
# Function to find the distinct pairs from
# 1-a & 1-b such that their sum is divisible by n.
def findCountOfPairs(a, b, n):
    ans = 0
    for i in range(1, a + 1):
 
        # For each integer from 1 to a
        # b/n integers exists such that pair
        # / sum is divisible by n
        ans += b//n
 
        # If (i%n +b%n ) >= n one more pair is possible
        ans += 1 if (i % n + b % n) >= n else 0
 
    return ans
 
# Driver code
a = 5; b = 13; n = 3
print(findCountOfPairs(a, b, n))
 
# This code is contributed by Shrikant13

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C#

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// C# program for the above approach
using System;
 
class GFG
{
     
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
static int findCountOfPairs(int a, int b, int n)
{
    int ans = 0;
 
    // Iterate over 1 to a to find distinct pairs
    for (int i = 1; i <= a; i++)
    {
        // For each integer from 1 to a
        // b/n integers exists such that pair
        // sum is divisible by n
        ans += b / n;
 
        // If (i%n +b%n ) >= n one more pair is possible
        ans += (i % n + b % n) >= n ? 1 : 0;
    }
 
    // Return answer
    return ans;
}
 
// Driver code
static public void Main ()
{
    int a = 5, b = 13, n = 3;
    Console.WriteLine(findCountOfPairs(a, b, n));
}
}
 
// This code has been contributed by ajit.

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PHP

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<?php
// PHP implementation of above approach
 
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
function findCountOfPairs($a, $b, $n)
{
    $ans = 0;
 
    // Iterate over 1 to a to find
    // distinct pairs
    for ($i = 1; $i <= $a; $i++)
    {
         
        // For each integer from 1 to a
        // b/n integers exists such that pair
        // sum is divisible by n
        $ans += (int)($b / $n);
 
        // If (i%n +b%n ) >= n one more
        // pair is possible
        $ans += (($i % $n ) +
                 ($b % $n)) >= $n ? 1 : 0;
    }
 
    // Return answer
    return $ans;
}
 
// Driver code
$a = 5;
$b = 13;
$n = 3;
echo findCountOfPairs($a, $b, $n);
 
// This code is contributed by akt_mit.
?>

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Output: 

22



 

Time complexity: O(N)



Second Approach:-  This Approach Is a little Tricky. Here we find how much pair for a multiple of  N.

First:- Keep (a<b), if not then make using swap.

Second:- Start a for loop from the lowest multiple of N and go through the multiple.

  • Now Smallest element ( a ) is greater than or equally current multiple then we add ((Current_Multiple) – 1) pair to the ans.
  • Now a is smaller But b is greater than or equally current multiple than we add a to the ans.
  • Now if a and b both smaller than we count the remaining pair for add a – (current_multiple – b ) + 1.
  • Break The Loop.

Below is the implementation of the above logic.

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
int findCountOfPairs(int a, int b, int n)
{
    if (a > b)
    {
        // if first element is bigger then swap
        swap(a, b);
    }
    int temp = 1, count = 0;
    // count is store the number of pair.
    for (int i = n; temp > 0; i += n)
    {
        // we use temp for breking a loop.
        if (a >= i)
        {
            // count when a is greter.
            temp = i - 1;
        }
        else if (b >= i)
        {
            // Count when a is smaller but
            // b is grater
            temp = a;
        }
        else if (i > b)
        {
            // Count when a and b both are smaller
            temp = a - (i - b) + 1;
        }
        if (temp > 0) //breaking condition
        {
            // For storing The pair in count.
            count += temp;
        }
    }
    // return the number of pairs.
    return count;
}
 
// Driver code
int main()
{
    int a = 5, b = 13, n = 3;
    cout << findCountOfPairs(a, b, n);
 
    return 0;
}
// contribute by Vivek Javiya

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Java

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// Java implementation of
// the above approach
class GFG{
     
// Function to find the
// distinct pairs from
// 1-a & 1-b such that
// their sum is divisible by n.
public static int findCountOfPairs(int a,
                                   int b,
                                   int n)
{
  if (a > b)
  {
    // if first element is
    // bigger then swap
    int temp = a;
    a = b;
    b = temp;
  }
   
  int temp = 1, count = 0;
   
  // count is store the
  // number of pair.
  for (int i = n; temp > 0; i += n)
  {
    // we use temp for
    // breaking a loop.
    if (a >= i)
    {
      // count when a
      // is greter.
      temp = i - 1;
    }
    else if (b >= i)
    {
      // Count when a is
      // smaller but
      // b is grater
      temp = a;
    }
    else if (i > b)
    {
      // Count when a and b
      // both are smaller
      temp = a - (i - b) + 1;
    }
     
    //breaking condition
    if (temp > 0)
    {
      // For storing The
      // pair in count.
      count += temp;
    }
  }
   
  // return the number
  // of pairs.
  return count;
}
   
// Driver code
public static void main(String[] args)
{
  int a = 5, b = 13, n = 3;
  System.out.print(findCountOfPairs(a,
                                    b, n));
}
}
 
// This code is contributed by divyeshrabadiya07

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Output :

22

Efficient Approach: For solving the problem efficiently break it into four-part and solve as: 

  • Each integer from range 1 to N*(a/N) will have exactly b/N integers from 1 to N*(b/N) whose sum is divisible by N.
  • There exist a/N integers from range 1 to N*(a/N) which can form pairs with b%N integer ranging from N*(b/N) to b.
  • There exist a%N integers from range N*(a/N) to a which can form pairs with b/N integer ranging from 1 to N*(b/N).
  • There exists (a%N + b%N)/N integers from range N*(a/N) to a and from N*(b/N) to b which can form pair whose sum is divisible by N.

Below is the implementation of the above approach:
 

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
int findCountOfPairs(int a, int b, int n)
{
    int ans = 0;
 
    // pairs from 1 to n*(a/n) and 1 to n*(b/n)
    ans += n * (a / n) * (b / n);
 
    // pairs from 1 to n*(a/n) and n*(b/n) to b
    ans += (a / n) * (b % n);
 
    // pairs from n*(a/n) to a and 1 to n*(b/n)
    ans += (a % n) * (b / n);
 
    // pairs from n*(a/n) to a and  n*(b/n) to b
    ans += ((a % n) + (b % n)) / n;
 
    // Return answer
    return ans;
}
 
// Driver code
int main()
{
    int a = 5, b = 13, n = 3;
    cout << findCountOfPairs(a, b, n);
 
    return 0;
}

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Java

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// Java implementation of above approach
import java.io.*;
 
class GFG
{
     
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
static int findCountOfPairs(int a, int b, int n)
{
    int ans = 0;
 
    // pairs from 1 to n*(a/n) and 1 to n*(b/n)
    ans += n * (a / n) * (b / n);
 
    // pairs from 1 to n*(a/n) and n*(b/n) to b
    ans += (a / n) * (b % n);
 
    // pairs from n*(a/n) to a and 1 to n*(b/n)
    ans += (a % n) * (b / n);
 
    // pairs from n*(a/n) to a and n*(b/n) to b
    ans += ((a % n) + (b % n)) / n;
 
    // Return answer
    return ans;
}
 
// Driver code
public static void main (String[] args)
{
    int a = 5, b = 13, n = 3;
    System.out.println (findCountOfPairs(a, b, n));
}
}
 
// This code is contributed by ajit..

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Python3

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# Python 3 implementation of above approach
 
# Function to find the distinct pairs from
# 1-a & 1-b such that their sum is divisible by n.
def findCountOfPairs(a, b, n):
    ans = 0
 
    # pairs from 1 to n*(a/n) and 1 to n*(b/n)
    ans += n * int(a / n) * int(b / n)
 
    # pairs from 1 to n*(a/n) and n*(b/n) to b
    ans += int(a / n) * (b % n)
 
    # pairs from n*(a/n) to a and 1 to n*(b/n)
    ans += (a % n) * int(b / n)
 
    # pairs from n*(a/n) to a and n*(b/n) to b
    ans += int(((a % n) + (b % n)) / n);
 
    # Return answer
    return ans
 
# Driver code
if __name__ == '__main__':
    a = 5
    b = 13
    n = 3
    print(findCountOfPairs(a, b, n))
 
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# implementation of above approach
using System;
 
class GFG
{
         
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
 
static int findCountOfPairs(int a, int b, int n)
{
    int ans = 0;
 
    // pairs from 1 to n*(a/n) and 1 to n*(b/n)
    ans += n * (a / n) * (b / n);
 
    // pairs from 1 to n*(a/n) and n*(b/n) to b
    ans += (a / n) * (b % n);
 
    // pairs from n*(a/n) to a and 1 to n*(b/n)
    ans += (a % n) * (b / n);
 
    // pairs from n*(a/n) to a and n*(b/n) to b
    ans += ((a % n) + (b % n)) / n;
 
    // Return answer
    return ans;
}
 
// Driver code
    static public void Main (){
    int a = 5, b = 13, n = 3;
    Console.WriteLine(findCountOfPairs(a, b, n));
}
}
 
// This code is contributed by @Tushil

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PHP

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<?php
// PHP implementation of above approach
 
// Function to find the distinct pairs from
// 1-a & 1-b such that their sum is divisible by n.
function findCountOfPairs($a, $b, $n)
{
    $ans = 0;
 
    // pairs from 1 to n*(a/n) and 1 to n*(b/n)
    $ans += $n * (int)($a / $n) * (int)($b / $n);
 
    // pairs from 1 to n*(a/n) and n*(b/n) to b
    $ans += (int)($a / $n) * ($b % $n);
 
    // pairs from n*(a/n) to a and 1 to n*(b/n)
    $ans += ($a % $n) * (int)($b / $n);
 
    // pairs from n*(a/n) to a and n*(b/n) to b
    $ans += (($a % $n) + (int)($b % $n)) / $n;
 
    // Return answer
    return $ans;
}
 
// Driver code
$a = 5;
$b = 13;
$n = 3;
echo findCountOfPairs($a, $b, $n);
 
// This code is contributed by ajit.
?>

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Output: 

22



 

Time complexity: O(1)

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