Given three positive integers a, b and n, our task is to find the total count of all the numbers K ranging from 0 to n which satisfies the given equation (( k % a ) % b) = (( k % b ) % a).
Examples:
Input: a = 3, b = 4, n = 25
Output: 10
Explanation:
The values which satisfies the above equation are 0 1 2 3 12 13 14 15 24 25. For example, for K = 13; ((13 % 3) % 4) gives 1 and ((13 % 4) % 3) also gives 1 as output.
Input: a = 1, b = 13, n = 500
Output: 501
Explanation:
In total there are 501 numbers between 0 and 500 which satisfies the given equation.
Approach:
To solve the problem mentioned above we have the given condition (( k % a ) % b) = (( k % b ) % a) which will always be satisfied for numbers from 0 to max(a, b) – 1. So according to the statement provided above if we have a <= b then check all number from 0 to b-1 and we have the following two cases:
- We calculate (k % a) % b, in this case answer will always be (k % a) since the value of (k % a) will always be less than b.
- We calculate (k % b) % a, in this case also answer will always be (k % a) because (k % b) will return k as k is less than b.
Similarly, we can check the cases for a > b. So now we need to check all numbers which are divisible by both a and b in the range 0 to n. This can be found with the help of LCM of a and b. So, now we can easily find the number of multiples of the LCM in the range 0 to n by diving n by LCM. We will add 1 to the multiples to include 0 as a multiple. And then we have to multiply the number of multiples by max(a, b) so that we can find all numbers which satisfy the given condition. But if the sum of the last multiple and max(a, b) exceeds our range of n numbers then we need to exclude the extra numbers.
Below is the implementation of the above approach:
C++
// C++ implementation to Find the total // count of all the numbers from 0 to n which// satisfies the given equation for a value K#include <bits/stdc++.h>using namespace std;// Function to find the valuesint findAns(int a, int b, int n){ // Calculate the LCM int lcm = (a * b) / __gcd(a, b); // Calculate the multiples of lcm int multiples = (n / lcm) + 1; // Find the values which satisfies // the given condition int answer = max(a, b) * multiples; // Subtract the extra values int lastvalue = lcm * (n / lcm) + max(a, b); if (lastvalue > n) answer = answer - (lastvalue - n - 1); // Return the final result return answer;}// Driver codeint main(){ int a = 1, b = 13, n = 500; cout << findAns(a, b, n) << endl;} |
Java
// Java implementation to find the total // count of all the numbers from 0 to n which// satisfies the given equation for a value Kclass GFG{// Function to find the valuesstatic int findAns(int a, int b, int n){ // Calculate the LCM int lcm = (a * b) / __gcd(a, b); // Calculate the multiples of lcm int multiples = (n / lcm) + 1; // Find the values which satisfies // the given condition int answer = Math.max(a, b) * multiples; // Subtract the extra values int lastvalue = lcm * (n / lcm) + Math.max(a, b); if (lastvalue > n) answer = answer - (lastvalue - n - 1); // Return the final result return answer;}static int __gcd(int a, int b) { return b == 0 ? a : __gcd(b, a % b); } // Driver codepublic static void main(String[] args){ int a = 1, b = 13, n = 500; System.out.print(findAns(a, b, n) + "\n");}}// This code is contributed by Amit Katiyar |
Python3
# Python3 implementation to find the total # count of all the numbers from 0 to n which# satisfies the given equation for a value K# Function to find the valuesdef findAns(a, b, n): # Calculate the LCM lcm = (a * b) // __gcd(a, b); # Calculate the multiples of lcm multiples = (n // lcm) + 1; # Find the values which satisfies # the given condition answer = max(a, b) * multiples; # Subtract the extra values lastvalue = lcm * (n // lcm) + max(a, b); if (lastvalue > n): answer = answer - (lastvalue - n - 1); # Return the final result return answer;def __gcd(a, b): if(b == 0): return a; else: return __gcd(b, a % b); # Driver codeif __name__ == '__main__': a = 1; b = 13; n = 500; print(findAns(a, b, n)); # This code is contributed by 29AjayKumar |
C#
// C# implementation to find the total // count of all the numbers from 0 to n which// satisfies the given equation for a value Kusing System;class GFG{// Function to find the valuesstatic int findAns(int a, int b, int n){ // Calculate the LCM int lcm = (a * b) / __gcd(a, b); // Calculate the multiples of lcm int multiples = (n / lcm) + 1; // Find the values which satisfies // the given condition int answer = Math.Max(a, b) * multiples; // Subtract the extra values int lastvalue = lcm * (n / lcm) + Math.Max(a, b); if (lastvalue > n) { answer = answer - (lastvalue - n - 1); } // Return the readonly result return answer;}static int __gcd(int a, int b) { return b == 0 ? a : __gcd(b, a % b); } // Driver codepublic static void Main(String[] args){ int a = 1, b = 13, n = 500; Console.Write(findAns(a, b, n) + "\n");}}// This code is contributed by sapnasingh4991 |
Javascript
<script>// javascript implementation to find the total // count of all the numbers from 0 to n which// satisfies the given equation for a value K // Function to find the values function findAns(a , b , n) { // Calculate the LCM var lcm = (a * b) / __gcd(a, b); // Calculate the multiples of lcm var multiples = (n / lcm) + 1; // Find the values which satisfies // the given condition var answer = Math.max(a, b) * multiples; // Subtract the extra values var lastvalue = lcm * (n / lcm) + Math.max(a, b); if (lastvalue > n) answer = answer - (lastvalue - n - 1); // Return the final result return answer; } function __gcd(a , b) { return b == 0 ? a : __gcd(b, a % b); } // Driver code var a = 1, b = 13, n = 500; document.write(findAns(a, b, n) + "\n");// This code contributed by Rajput-Ji </script> |
501
Time Complexity: O(log(min(a, b)), for using __gcd(a,b) method the overall time complexity is O(log(min(a,b))).
Auxiliary Space: O(1) because constant extra space is required.
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