Given two integer N and K, the task is to find the sum of all the numbers from the range [1, N] excluding those which are powers of K.
Examples:
Input: N = 10, K = 3
Output: 42
2 + 4 + 5 + 6 + 7 + 8 + 10 = 42
1, 3 and 9 are excluded as they are powers of 3.Input: N = 200, K = 30
Output: 20069
Approach: Find the sum of the following series:
- pwrK: The sum of all the powers of K from [1, N] i.e. K0 + K1 + K2 + … + Kr such that Kr ≤ N
- sumAll: The sum of all the integers from the range [1, N] i.e. (N * (N + 1)) / 2.
The result will be sumAll – pwrK
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; #define ll long long int // Function to return the sum of all the // powers of k from the range [1, n] ll sumPowersK(ll n, ll k) { // To store the sum of the series ll sum = 0, num = 1; // While current power of k <= n while (num <= n) { // Add current power to the sum sum += num; // Next power of k num *= k; } // Return the sum of the series return sum; } // Find to return the sum of the // elements from the range [1, n] // excluding those which are powers of k ll getSum(ll n, ll k) { // Sum of all the powers of k from [1, n] ll pwrK = sumPowersK(n, k); // Sum of all the elements from [1, n] ll sumAll = (n * (n + 1)) / 2; // Return the required sum return (sumAll - pwrK); } // Driver code int main() { ll n = 10, k = 3; cout << getSum(n, k); return 0; } |
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Java
// Java implementation of the approach import java.io.*; class GFG { // Function to return the sum of all the // powers of k from the range [1, n] static long sumPowersK(long n, long k) { // To store the sum of the series long sum = 0, num = 1; // While current power of k <= n while (num <= n) { // Add current power to the sum sum += num; // Next power of k num *= k; } // Return the sum of the series return sum; } // Find to return the sum of the // elements from the range [1, n] // excluding those which are powers of k static long getSum(long n, long k) { // Sum of all the powers of k from [1, n] long pwrK = sumPowersK(n, k); // Sum of all the elements from [1, n] long sumAll = (n * (n + 1)) / 2; // Return the required sum return (sumAll - pwrK); } // Driver code public static void main (String[] args) { long n = 10, k = 3; System.out.println( getSum(n, k)); } } // This code is contributed by anuj_67.. |
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Python3
# Python3 implementation of the approach # Function to return the sum of all the # powers of k from the range [1, n] def sumPowersK(n, k) : # To store the sum of the series sum = 0; num = 1; # While current power of k <= n while (num <= n) : # Add current power to the sum sum += num; # Next power of k num *= k; # Return the sum of the series return sum; # Find to return the sum of the # elements from the range [1, n] # excluding those which are powers of k def getSum(n, k) : # Sum of all the powers of k from [1, n] pwrK = sumPowersK(n, k); # Sum of all the elements from [1, n] sumAll = (n * (n + 1)) / 2; # Return the required sum return (sumAll - pwrK); # Driver code if __name__ == "__main__" : n = 10; k = 3; print(getSum(n, k)); # This code is contributed by AnkitRai01 |
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C#
// C# implementation of the approach using System; class GFG { // Function to return the sum of all the // powers of k from the range [1, n] static long sumPowersK(long n, long k) { // To store the sum of the series long sum = 0, num = 1; // While current power of k <= n while (num <= n) { // Add current power to the sum sum += num; // Next power of k num *= k; } // Return the sum of the series return sum; } // Find to return the sum of the // elements from the range [1, n] // excluding those which are powers of k static long getSum(long n, long k) { // Sum of all the powers of k from [1, n] long pwrK = sumPowersK(n, k); // Sum of all the elements from [1, n] long sumAll = (n * (n + 1)) / 2; // Return the required sum return (sumAll - pwrK); } // Driver code public static void Main () { long n = 10, k = 3; Console.WriteLine( getSum(n, k)); } } // This code is contributed by anuj_67.. |
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Output:
42
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