前N个自然数之和，不是K的幂

给定两个整数$n$和$k$ ，任务是找到[1，n]范围内的所有数字之和，不包括k的正幂数，即k，k ² ，k ³等。
例子：

Input: n = 10, k = 3 
Output: 43 
1 + 2 + 4 + 5 + 6 + 7 + 8 + 10 = 43 
3 and 9 are excluded as they are powers of 3
Input: n = 11, k = 2 
Output: 52 
 

方法：

• 存储第一个的总和$n$变量中的自然数$sum$即sum =(n *(n + 1))/ 2 。
• 现在，对于……的每一个积极力量$k$小于$n$ ，从中减去$sum$ 。
• 打印变量的值$sum$到底。

下面是上述方法的实现：

C++

// C++ program to find the sum of
// first n natural numbers which are
// not positive powers of k
#include 
using namespace std;
 
// Function to return the sum of
// first n natural numbers which are
// not positive powers of k
int find_sum(int n, int k)
{
    // sum of first n natural numbers
    int total_sum = (n * (n + 1)) / 2;
 
    int power = k;
    while (power <= n) {
 
        // subtract all positive powers
        // of k which are less than n
        total_sum -= power;
 
        // next power of k
        power *= k;
    }
 
    return total_sum;
}
 
// Driver code
int main()
{
    int n = 11, k = 2;
 
    cout << find_sum(n, k);
 
    return 0;
}

Java

// Java program to find the sum of
// first n natural numbers which are
// not positive powers of k
import java.io.*;
 
class GFG {
   
 
 
// Function to return the sum of
// first n natural numbers which are
// not positive powers of k
static int find_sum(int n, int k)
{
    // sum of first n natural numbers
    int total_sum = (n * (n + 1)) / 2;
 
    int power = k;
    while (power <= n) {
 
        // subtract all positive powers
        // of k which are less than n
        total_sum -= power;
 
        // next power of k
        power *= k;
    }
 
    return total_sum;
}
 
// Driver code
 
    public static void main (String[] args) {
        int n = 11, k = 2;
 
    System.out.println(find_sum(n, k));
    }
}
// This code is contributed by inder_verma..

Python3

# Python 3 program to find the sum of
# first n natural numbers which are
# not positive powers of k
 
# Function to return the sum of
# first n natural numbers which are
# not positive powers of k
def find_sum(n, k):
 
    # sum of first n natural numbers
    total_sum = (n * (n + 1)) // 2
    power = k
    while power <= n:
 
        # subtract all positive powers
        # of k which are less than n
        total_sum -= power
 
        # next power of k
        power *= k
    return total_sum
 
# Driver code
n = 11; k = 2
print(find_sum(n, k))
 
# This code is contributed
# by Shrikant13

C#

// C# program to find the sum of
// first n natural numbers which are
// not positive powers of k
using System;
 
class GFG {
 
 
 
// Function to return the sum of
// first n natural numbers which are
// not positive powers of k
static int find_sum(int n, int k)
{
    // sum of first n natural numbers
    int total_sum = (n * (n + 1)) / 2;
 
    int power = k;
    while (power <= n) {
 
        // subtract all positive powers
        // of k which are less than n
        total_sum -= power;
 
        // next power of k
        power *= k;
    }
 
    return total_sum;
}
 
// Driver code
 
    public static void Main () {
        int n = 11, k = 2;
 
    Console.WriteLine(find_sum(n, k));
    }
}
// This code is contributed by inder_verma..

Javascript

 
// Javascript program to find the sum of
// first n natural numbers which are
// not positive powers of k
 
// Function to return the sum of
// first n natural numbers which are
// not positive powers of k
function find_sum(n, k)
{
    // sum of first n natural numbers
    let total_sum = (n * (n + 1)) / 2;
 
    let power = k;
    while (power <= n) {
 
        // subtract all positive powers
        // of k which are less than n
        total_sum -= power;
 
        // next power of k
        power *= k;
    }
 
    return total_sum;
}
 
// Driver code
    let n = 11, k = 2;
 
    document.write(find_sum(n, k));
     
// This code is contributed by Mayank Tyagi
 

输出：

52