找出1到n的数字之和，不包括K的幂

给定两个整数N和K ，任务是从范围[1，N]中查找所有数字的和，不包括K的幂。

例子：

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

方法：找到以下系列的总和：

1. pwrK：K个的所有权力从[1，N]即^ķ0 + K ¹ + K ² + … + K ^R，如^的Kr≤n中的总和
2. sumAll：范围为[1，N]的所有整数的总和，即(N *(N +1))/ 2 。

结果将是sumAll – pwrK

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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;
}

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..

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

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..

输出：

42