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Count numbers less than N whose modulo with A is equal to B

  • Last Updated : 05 Jul, 2021

Given three non-negative integers A, B, and N where A is non-zero, the task is to find the number of integers less than or equal to N whose modulo with A gives the value B.

Examples:

Input: A = 6, B = 3, N = 15
Output: 3
Explanation: The numbers 3, 9 and 15 are less than or equal to N (= 15) and their modulo with A (= 6) is equal to B (= 3). Therefore, the count such numbers is 3.

Input: A = 4, B = 1, C = 8
Output: 2

 

Approach: The given problem can be solved by using an observation based on mathematics. Follow the steps below to solve the problem:



  • If the value of B is at least A, then print 0, as there can’t be any such numbers whose modulo with A gives the value B.
  • Otherwise, If the value of B is 0, then print the value of C / A as the count of such numbers whose modulo with A gives the value B.
  • Otherwise, perform the following steps:
    • Initialize a variable, say ans, with the floor value of C / A.
    • If the value of (ans * A + B) is less than or equal to N, then increment the value of ans by 1.
    • After completing the above steps, print the value of ans as the count of such numbers whose modulo with A gives the value B.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count numbers less than
// N, whose modulo with A gives B
void countValues(int A, int B, int C)
{
    // If the value of B at least A
    if (B >= A) {
        cout << 0;
        return;
    }
 
    // If the value of B is 0 or not
    if (B == 0) {
        cout << C / A;
        return;
    }
 
    // Stores the resultant count
    // of numbers less than N
    int ans = C / A;
 
    if (ans * A + B <= C) {
 
        // Update the value of ans
        ans++;
    }
 
    // Print the value of ans
    cout << ans;
}
 
// Driver Code
int main()
{
    int A = 6, B = 3, N = 15;
    countValues(A, B, N);
 
    return 0;
}

Java




// Java program for the above approach
public class MyClass
{
     
// Function to count numbers less than
// N, whose modulo with A gives B
static void countValues(int A, int B, int C)
{
   
    // If the value of B at least A
    if (B >= A) {
        System.out.println(0);
        return;
    }
 
    // If the value of B is 0 or not
    if (B == 0) {
        System.out.println(C / A);
        return;
    }
 
    // Stores the resultant count
    // of numbers less than N
    int ans = C / A;
 
    if (ans * A + B <= C) {
 
        // Update the value of ans
        ans++;
    }
 
    // Print the value of ans
    System.out.println(ans);
}
 
// Driver Code
public static void main(String args[])
{
    int A = 6, B = 3, N = 15;
    countValues(A, B, N);
 
}  
}
 
// This code in contributed by SoumikMondal

Python3




# Python3 program for the above approach
 
# Function to count numbers less than
# N, whose modulo with A gives B
def countValues(A, B, C):
     
    # If the value of B at least A
    if (B >= A):
        print(0)
        return
 
    # If the value of B is 0 or not
    if (B == 0):
        print(C // A)
        return
 
    # Stores the resultant count
    # of numbers less than N
    ans = C//A
 
    if (ans * A + B <= C):
 
        # Update the value of ans
        ans += 1
 
    # Print the value of ans
    print(ans)
 
# Driver Code
if __name__ == '__main__':
     
    A = 6
    B = 3
    N = 15
     
    countValues(A, B, N)
 
# This code is contributed by SURENDRA_GANGWAR

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to count numbers less than
// N, whose modulo with A gives B
static void countValues(int A, int B, int C)
{
     
    // If the value of B at least A
    if (B >= A)
    {
        Console.Write(0);
        return;
    }
 
    // If the value of B is 0 or not
    if (B == 0)
    {
        Console.Write(C / A);
        return;
    }
 
    // Stores the resultant count
    // of numbers less than N
    int ans = C / A;
 
    if (ans * A + B <= C)
    {
         
        // Update the value of ans
        ans++;
    }
 
    // Print the value of ans
    Console.Write(ans);
}
 
// Driver code
public static void Main()
{
    int A = 6, B = 3, N = 15;
    countValues(A, B, N);
}
}
 
// This code is contributed by sanjoy_62

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to count numbers less than
// N, whose modulo with A gives B
function countValues(A, B, C)
{
    // If the value of B at least A
    if (B >= A) {
        document.write(0);
        return;
    }
 
    // If the value of B is 0 or not
    if (B == 0) {
        document.write(parseInt(C / A));
        return;
    }
 
    // Stores the resultant count
    // of numbers less than N
    let ans = parseInt(C / A);
 
    if (ans * A + B <= C) {
 
        // Update the value of ans
        ans++;
    }
 
    // Print the value of ans
    document.write(ans);
}
 
// Driver Code
    let A = 6, B = 3, N = 15;
    countValues(A, B, N);
 
</script>
Output: 
3

 

Time Complexity: O(1)
Auxiliary Space: O(1)

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