不能被N整除的前K个数字的总和

给定两个数字N和K ，任务是找到不能被N整除的前K个数字的总和。

例子：

Input: N = 5, K = 10
Output: 63
Explanation: Sum of { 1, 2, 3, 4, 6, 7, 8, 9, 11, 12 } is 63.

Input: N = 3, k = 13
Output: 127
Explanation: Sum of { 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19 } is 127.

方法：为了解决此问题，我们需要遵循以下步骤：

• 用(K /(N – 1))* N计算N的最后一个倍数
• 计算值K％(N – 1) 。如果余数为0，则(K /(N – 1))* N – 1给出不能被N整除的最后一个值。否则，我们需要将余数加到(K /(N – 1))* N 。
• 计算总和，直到最后一个值，该最后一个值不能被上述步骤中获得的N整除。
• 计算N的倍数之和，然后从上一步中计算出的总和中减去以获得所需的结果。

下面的代码是上述方法的实现：

C++

// C++ Program to calculate
// the sum of first K
// numbers not divisible by N
  
#include 
using namespace std;
  
// Function to find the sum
int findSum(int n, int k)
{
    // Find the last multiple of N
    int val = (k / (n - 1)) * n;
  
    int rem = k % (n - 1);
  
    // Find the K-th non-multiple of N
    if (rem == 0) {
        val = val - 1;
    }
    else {
        val = val + rem;
    }
  
    // Calculate the  sum of
    // all elements from 1 to val
    int sum = (val * (val + 1)) / 2;
  
    // Calculate the sum of
    // all multiples of N
    // between 1 to val
    int x = k / (n - 1);
    int sum_of_multiples
        = (x
           * (x + 1) * n)
          / 2;
    sum -= sum_of_multiples;
  
    return sum;
}
  
// Driver code
int main()
{
    int n = 7, k = 13;
    cout << findSum(n, k)
         << endl;
}

Java

// Java program to calculate 
// the sum of first K numbers
// not divisible by N
import java.util.*;
  
class GFG {
  
// Function to find the sum
static int findSum(int n, int k)
{
      
    // Find the last multiple of N
    int val = (k / (n - 1)) * n;
  
    int rem = k % (n - 1);
  
    // Find the K-th non-multiple of N
    if (rem == 0)
    {
        val = val - 1;
    }
    else 
    {
        val = val + rem;
    }
  
    // Calculate the sum of
    // all elements from 1 to val
    int sum = (val * (val + 1)) / 2;
  
    // Calculate the sum of
    // all multiples of N
    // between 1 to val
    int x = k / (n - 1);
    int sum_of_multiples = (x * (x + 1) * n) / 2;
    sum -= sum_of_multiples;
  
    return sum;
}
  
// Driver code
public static void main(String[] args)
{
    int n = 7, k = 13;
  
    System.out.println(findSum(n, k));
}
}
  
// This code is contributed by offbeat

Python3

# Python3 Program to calculate
# the sum of first K
# numbers not divisible by N
  
# Function to find the sum
def findSum(n, k):
  
    # Find the last multiple of N
    val = (k // (n - 1)) * n;
  
    rem = k % (n - 1);
  
    # Find the K-th non-multiple of N
    if (rem == 0):
        val = val - 1;
      
    else:
        val = val + rem;
      
    # Calculate the sum of
    # all elements from 1 to val
    sum = (val * (val + 1)) // 2;
  
    # Calculate the sum of
    # all multiples of N
    # between 1 to val
    x = k // (n - 1);
    sum_of_multiples = (x * (x + 1) * n) // 2;
    sum -= sum_of_multiples;
  
    return sum;
  
# Driver code
n = 7; k = 13;
print(findSum(n, k))
  
# This code is contributed by Code_Mech

C#

// C# program to calculate 
// the sum of first K numbers
// not divisible by N
using System;
  
class GFG{
  
// Function to find the sum
static int findSum(int n, int k)
{
      
    // Find the last multiple of N
    int val = (k / (n - 1)) * n;
  
    int rem = k % (n - 1);
  
    // Find the K-th non-multiple of N
    if (rem == 0)
    {
        val = val - 1;
    }
    else
    {
        val = val + rem;
    }
  
    // Calculate the sum of
    // all elements from 1 to val
    int sum = (val * (val + 1)) / 2;
  
    // Calculate the sum of
    // all multiples of N
    // between 1 to val
    int x = k / (n - 1);
    int sum_of_multiples = (x * (x + 1) * n) / 2;
    sum -= sum_of_multiples;
  
    return sum;
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 7, k = 13;
  
    Console.WriteLine(findSum(n, k));
}
}
  
// This code is contributed by 29AjayKumar

输出：

99

时间复杂度： O(1)
辅助空间复杂度： O(1)