Count of N digit 每K个连续数字之和相等的数| 2套

给定两个整数N和K ，任务是找到所有可能的N 位数字的计数，这些数字的每K个连续数字的总和都相等。
例子：

Input: N = 2, K=1 
Output: 9 
Explanation: 
All two digit numbers satisfying the required conditions are {11, 22, 33, 44, 55, 66, 77, 88, 99}
Input: N = 3, K = 2 
Output: 90 
 

朴素和滑动窗口方法：对于最简单的方法和基于滑动窗口技术的方法，请参阅 N 位数字的计数，其每 K 个连续数字的总和相等。
对数法：
为了使K 个连续元素的总和始终相等，整个数字由其前K位数字控制。

• 第i^位数字将等于第(ik)^位数字，以满足条件，即每K个连续数字的总和相同。

Illustration: 
N = 5 and K = 2 
If the first two digits are 1 and 2, then the number has to be 12121 so that sum of every 2 consecutive digits is 3. 
Observe that the first 2 digits i.e. the first K digits repeats itself. 
 

因此，为了解决这个问题，现在的任务是找出等于10 ^K – 10 ^(K-1)的K 位数字的总数。因此，打印计算值作为答案。
下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to count the number of
// N-digit numbers such that sum of
// every K consecutive digits are equal
void count(int n, int k)
{
    long count = (long)(pow(10, k) - pow(10, k - 1));
 
    // Print the answer
    cout << (count);
}
 
// Driver Code
int main()
{
    int n = 2, k = 1;
    count(n, k);
}
 
// This code is contributed by Ritik Bansal

Java

// Java Program to implement
// the above approach
class GFG {
 
    // Function to count the number of
    // N-digit numbers such that sum of
    // every K consecutive digits are equal
    public static void count(int n, int k)
    {
        long count
            = (long)(Math.pow(10, k)
                     - Math.pow(10, k - 1));
 
        // Print the answer
        System.out.print(count);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 2, k = 1;
        count(n, k);
    }
}

Python3

# Python3 program to implement
# the above approach
 
# Function to count the number of
# N-digit numbers such that sum of
# every K consecutive digits are equal
def count(n, k):
     
    count = (pow(10, k) - pow(10, k - 1));
 
    # Print the answer
    print(count);
     
# Driver Code
if __name__ == '__main__':
     
    n = 2;
    k = 1;
     
    count(n, k);
     
# This code is contributed by 29AjayKumar

C#

// C# Program to implement
// the above approach
using System;
class GFG{
   
  // Function to count the number of
  // N-digit numbers such that sum of
  // every K consecutive digits are equal
  public static void count(int n, int k)
  {
    long count = (long)(Math.Pow(10, k) -
                        Math.Pow(10, k - 1));
 
    // Print the answer
    Console.Write(count);
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int n = 2, k = 1;
    count(n, k);
  }
}
 
// This code is contributed by Rohit_ranjan

Javascript

// Javascript Program to implement
// the above approach
 
// Function to count the number of
// N-digit numbers such that sum of
// every K consecutive digits are equal
function count(n, k)
{
    let count = Math.pow(10, k) - Math.pow(10, k - 1);
 
    // Print the answer
    document.write(count);
}
 
// Driver Code
let n = 2, k = 1;
count(n, k);
 
// This code is contributed by souravmahato348.

输出：

9

时间复杂度： O(log K)
辅助空间： O(1)