Given three integers N, A, B, the task is to find the remainder when the sum of integers which are divisible by either A or B in the range [1, N] is divided by the number of integers in this range.
Note: The numbers A and B are co-primes.
Examples:
Input: N = 88, A = 11, B = 8
Output: 8
Explanation:
There are a total of 18 numbers in the range [1, 88] which are divisible by either 8 or 11. They are:
{ 8, 11, 16, 22, 24, 32, 33, 40, 44, 48, 55, 56, 64, 66, 72, 77, 80, 88 }. Therefore, the sum of these numbers is 836. Therefore, 836 % 18 = 8.
Input: N = 100, A = 7, B = 19
Output: 13
Explanation:
There are a total of 19 numbers in the range [1, 100] which are divisible by either 7 or 19. They are:
{ 7, 14, 19, 21, 28, 35, 38, 42, 49, 56, 57, 63, 70, 76, 77, 84, 91, 95, 98 }. Therefore, the sum of these numbers is 1020. Therefore, 1020 % 19 = 13.
Naive Approach: The naive approach is to run a loop from 1 to N and count all the numbers which are divisible by either A or B while simultaneously adding those numbers in a variable to find its sum.
Time Complexity: O(N)
Efficient Approach: Efficient approach is to use the division method.
- By using the division method, the count of the numbers which are divisible either by A or B can be found in the constant time. The idea is to:
- Divide N by A to get the count of numbers divisible by A in the range [1, N].
- Divide N by B to get the count of numbers divisible by B in the range [1, N].
- Divide N by A * B to get the count of numbers divisible by both A and B.
- Add the values obtained in step 1 and step 2 and subtract the value obtained in step 3 to remove the numbers which have been counted twice.
- Since we are even interested in finding the numbers which are divisible in this range, the idea is to reduce the number of times the conditions are checked by the following way:
- Instead of completely relying on one loop, we can use two loops.
- One loop is to find the numbers divisible by A. Instead of incrementing the values by 1, we start the loop from A and increment it by A. This reduces the number of comparisons drastically.
- Similarly, another loop is used to find the numbers divisible by B.
- Since again, there might be repetitions in the numbers, the numbers are stored in a set so that there are no repetitions.
- Once the count and sum of the numbers are found, then directly modulo operation can be applied to compute the final answer.
Below is the implementation of the above approach:
CPP
// C++ implementation of the above approach#include <algorithm>#include <iostream>#include <set>#define ll long longusing namespace std;// Function to return the count of numbers// which are divisible by both A and B in// the range [1, N] in constant timell int countOfNum(ll int n, ll int a, ll int b){ ll int cnt_of_a, cnt_of_b, cnt_of_ab, sum; // Compute the count of numbers divisible by // A in the range [1, N] cnt_of_a = n / a; // Compute the count of numbers divisible by // B in the range [1, N] cnt_of_b = n / b; // Adding the counts which are // divisible by A and B sum = cnt_of_b + cnt_of_a; // The above value might contain repeated // values which are divisible by both // A and B. Therefore, the count of numbers // which are divisible by both A and B are found cnt_of_ab = n / (a * b); // The count computed above is subtracted to // compute the final count sum = sum - cnt_of_ab; return sum;}// Function to return the sum of numbers// which are divisible by both A and B// in the range [1, N]ll int sumOfNum(ll int n, ll int a, ll int b){ ll int i; ll int sum = 0; // Set to store the numbers so that the // numbers are not repeated set<ll int> ans; // For loop to find the numbers // which are divisible by A and insert // them into the set for (i = a; i <= n; i = i + a) { ans.insert(i); } // For loop to find the numbers // which are divisible by A and insert // them into the set for (i = b; i <= n; i = i + b) { ans.insert(i); } // For loop to iterate through the set // and find the sum for (auto it = ans.begin(); it != ans.end(); it++) { sum = sum + *it; } return sum;}// Driver codeint main(){ ll int N = 88; ll int A = 11; ll int B = 8; ll int count = countOfNum(N, A, B); ll int sumofnum = sumOfNum(N, A, B); cout << sumofnum % count << endl; return 0;} |
Java
// Java implementation of the above approachimport java.util.*;// Function to return the count of numbers// which are divisible by both A and B in// the range [1, N] in constant timeclass GFG { static int countOfNum( int n, int a, int b) { int cnt_of_a, cnt_of_b, cnt_of_ab, sum; // Compute the count of numbers divisible by // A in the range [1, N] cnt_of_a = n / a; // Compute the count of numbers divisible by // B in the range [1, N] cnt_of_b = n / b; // Adding the counts which are // divisible by A and B sum = cnt_of_b + cnt_of_a; // The above value might contain repeated // values which are divisible by both // A and B. Therefore, the count of numbers // which are divisible by both A and B are found cnt_of_ab = n / (a * b); // The count computed above is subtracted to // compute the final count sum = sum - cnt_of_ab; return sum; } // Function to return the sum of numbers // which are divisible by both A and B // in the range [1, N] static int sumOfNum( int n, int a, int b) { int i; int sum = 0; // Set to store the numbers so that the // numbers are not repeated Set< Integer> ans = new HashSet<Integer>(); // For loop to find the numbers // which are divisible by A and insert // them into the set for (i = a; i <= n; i = i + a) { ans.add(i); } // For loop to find the numbers // which are divisible by A and insert // them into the set for (i = b; i <= n; i = i + b) { ans.add(i); } // For loop to iterate through the set // and find the sum for (Integer it : ans) { sum = sum + it; } return sum; } // Driver code public static void main (String []args) { int N = 88; int A = 11; int B = 8; int count = countOfNum(N, A, B); int sumofnum = sumOfNum(N, A, B); System.out.print(sumofnum % count); } }// This code is contributed by chitranayal |
Python3
# Python3 implementation of the above approach# Function to return the count of numbers# which are divisible by both A and B in# the range [1, N] in constant timedef countOfNum(n, a, b): cnt_of_a, cnt_of_b, cnt_of_ab, sum = 0, 0, 0, 0 # Compute the count of numbers divisible by # A in the range [1, N] cnt_of_a = n // a # Compute the count of numbers divisible by # B in the range [1, N] cnt_of_b = n // b # Adding the counts which are # divisible by A and B sum = cnt_of_b + cnt_of_a # The above value might contain repeated # values which are divisible by both # A and B. Therefore, the count of numbers # which are divisible by both A and B are found cnt_of_ab = n // (a * b) # The count computed above is subtracted to # compute the final count sum = sum - cnt_of_ab return sum# Function to return the sum of numbers# which are divisible by both A and B# in the range [1, N]def sumOfNum(n, a, b): i = 0 sum = 0 # Set to store the numbers so that the # numbers are not repeated ans = dict() # For loop to find the numbers # which are divisible by A and insert # them into the set for i in range(a, n + 1, a): ans[i] = 1 # For loop to find the numbers # which are divisible by A and insert # them into the set for i in range(b, n + 1, b): ans[i] = 1 # For loop to iterate through the set # and find the sum for it in ans: sum = sum + it return sum# Driver codeif __name__ == '__main__': N = 88 A = 11 B = 8 count = countOfNum(N, A, B) sumofnum = sumOfNum(N, A, B) print(sumofnum % count)# This code is contributed by mohit kumar 29 |
C#
// C# implementation of the above approachusing System;using System.Collections.Generic;class GFG { // Function to return the count of numbers // which are divisible by both A and B in // the range [1, N] in constant time static int countOfNum( int n, int a, int b) { int cnt_of_a, cnt_of_b, cnt_of_ab, sum; // Compute the count of numbers divisible by // A in the range [1, N] cnt_of_a = n / a; // Compute the count of numbers divisible by // B in the range [1, N] cnt_of_b = n / b; // Adding the counts which are // divisible by A and B sum = cnt_of_b + cnt_of_a; // The above value might contain repeated // values which are divisible by both // A and B. Therefore, the count of numbers // which are divisible by both A and B are found cnt_of_ab = n / (a * b); // The count computed above is subtracted to // compute the readonly count sum = sum - cnt_of_ab; return sum; } // Function to return the sum of numbers // which are divisible by both A and B // in the range [1, N] static int sumOfNum( int n, int a, int b) { int i; int sum = 0; // Set to store the numbers so that the // numbers are not repeated HashSet< int> ans = new HashSet<int>(); // For loop to find the numbers // which are divisible by A and insert // them into the set for (i = a; i <= n; i = i + a) { ans.Add(i); } // For loop to find the numbers // which are divisible by A and insert // them into the set for (i = b; i <= n; i = i + b) { ans.Add(i); } // For loop to iterate through the set // and find the sum foreach (int it in ans) { sum = sum + it; } return sum; } // Driver code public static void Main(String []args) { int N = 88; int A = 11; int B = 8; int count = countOfNum(N, A, B); int sumofnum = sumOfNum(N, A, B); Console.Write(sumofnum % count); } }// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation of the above approach // Function to return the count of numbers // which are divisible by both A and B in // the range [1, N] in constant time function countOfNum(n,a,b) { let cnt_of_a, cnt_of_b, cnt_of_ab, sum; // Compute the count of numbers divisible by // A in the range [1, N] cnt_of_a = Math.floor(n / a); // Compute the count of numbers divisible by // B in the range [1, N] cnt_of_b = Math.floor(n / b); // Adding the counts which are // divisible by A and B sum = cnt_of_b + cnt_of_a; // The above value might contain repeated // values which are divisible by both // A and B. Therefore, the count of numbers // which are divisible by both A and B are found cnt_of_ab = Math.floor(n / (a * b)); // The count computed above is subtracted to // compute the final count sum = sum - cnt_of_ab; return sum; } // Function to return the sum of numbers // which are divisible by both A and B // in the range [1, N] function sumOfNum(n,a,b) { let i; let sum = 0; // Set to store the numbers so that the // numbers are not repeated let ans = new Set(); // For loop to find the numbers // which are divisible by A and insert // them into the set for (i = a; i <= n; i = i + a) { ans.add(i); } // For loop to find the numbers // which are divisible by A and insert // them into the set for (i = b; i <= n; i = i + b) { ans.add(i); } // For loop to iterate through the set // and find the sum for (let it of ans.values()) { sum = sum + it; } return sum; } // Driver code let N = 88; let A = 11; let B = 8; let count = countOfNum(N, A, B); let sumofnum = sumOfNum(N, A, B); document.write(sumofnum % count); // This code is contributed by rag2127</script> |
Output:
8
Time Complexity Analysis:
- The time taken to run the for loop to find the numbers which are divisible by A is O(N / A).
- The time taken to run the for loop to find the numbers which are divisible by B is O(N / B).
- Therefore, overall time complexity is O(N / A) + O(N / B).
Auxiliary Space: O(N/A + N/B)
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