计算所有N位数为X的倍数的数字

给定两个整数N和X ，任务是查找所有可能的N位数字的计数，其每一位数字都是X的倍数。

例子 ：

Input: N = 1, X = 3 
Output: 4 
Explanation: 
The single-digit numbers whose digits are multiple of 3 are 0, 3, 6 and 9. 
So the answer will be 4.
Input: N = 2, X = 4 
Output: 6 
Explanation: 
The two-digit numbers whose digits are multiple of 4 are 40, 44, 48, 80, 84, 88. 
So the answer will be 6.

天真的方法：最简单的方法是对所有N位数字进行迭代，即在[10 ^{N –} 1，10 ^N – 1]范围内，并针对每个数字检查数字的每个数字是否为X的倍数。增加此类数字的数量并打印最终数量。

时间复杂度： O((10 ^N – 10 ^{N – 1} )* N)
辅助空间： O(1)

高效方法：请按照以下步骤解决问题：

1. 计算X的倍数的总位数，将其表示为total_Multiples 。
2. 通过将上述所有数字排列在不同的位置以形成N位数字，可以获得所需的计数。

 
 

 Illustration: 
For example N = 2, X = 3 
Multiples of 3 are S = { 0, 3, 6, 9 } 

 

• To find all 2-digit numbers, arrange all the multiples of X to form a 2-digit number and count it.
• Arrange the numbers from the set S. Taking 0 for the most significant bit will result in 1-digit number, so there are 3 ways to arrange the most significant position of a number and there are 4 ways to arrange the last digit of a number whose digits are multiples of 3. 
  So total ways = 3*4 = 12 so the answer will be 12.

1. 从上图可以看出，如果M是X的所有倍数的计数，则所需的N个数字的计数(对于N> 1 )等于(M – 1)* M ^{N – 1} 。
2. 对于一位数字，答案是M。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
#define ll long long
 
// Function to calculate x^n
// using binary-exponentiation
ll power(ll x, ll n)
{
    // Stores the resultant power
    ll temp;
 
    if (n == 0)
        return 1;
 
    // Stores the value of x^(n/2)
    temp = power(x, n / 2);
    if (n % 2 == 0)
        return temp * temp;
    else
        return x * temp * temp;
}
 
// Function to count all N-digit numbers
// whose digits are multiples of x
ll count_Total_Numbers(ll n, ll x)
{
    ll total, multiples = 0;
 
    // Count all digits which
    // are multiples of x
    for (int i = 0; i < 10; i++) {
 
        // Check if current number
        // is a mutiple of X
        if (i % x == 0)
 
            // Increase count of multiples
            multiples++;
    }
 
    // Check if it's a 1 digit number
    if (n == 1)
        return multiples;
 
    // Count the total numbers
    total = (multiples - 1)
            * power(multiples, n - 1);
 
    // Return the total numbers
    return total;
}
 
// Driver Code
int main()
{
    // Given N and X
    ll N = 1, X = 3;
 
    // Function Call
    printf("%lld ",
           count_Total_Numbers(N, X));
 
    return 0;
}

Java

// Java program for
// the above approach
class GFG{
 
// Function to calculate x^n
// using binary-exponentiation
static int power(int x, int n)
{
   
  // Stores the resultant power
  int temp;
 
  if (n == 0)
    return 1;
 
  // Stores the value of x^(n/2)
  temp = power(x, n / 2);
  if (n % 2 == 0)
    return temp * temp;
  else
    return x * temp * temp;
}
 
// Function to count aint N-digit
// numbers whose digits are multiples
// of x
static int count_Total_Numbers(int n,
                               int x)
{
  int total, multiples = 0;
 
  // Count aint digits which
  // are multiples of x
  for (int i = 0; i < 10; i++)
  {
     
    // Check if current number
    // is a mutiple of X
    if (i % x == 0)
 
      // Increase count of multiples
      multiples++;
  }
 
  // Check if it's a 1 digit number
  if (n == 1)
    return multiples;
 
  // Count the total numbers
  total = (multiples - 1) *
           power(multiples, n - 1);
 
  // Return the total numbers
  return total;
}
 
// Driver Code
public static void main(String[] args)
{
  // Given N and X
  int N = 1, X = 3;
 
  // Function Call
  System.out.printf("%d ",
                    count_Total_Numbers(N, X));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program for the above approach
 
# Function to calculate x^n
# using binary-exponentiation
def power(x, n):
     
    # Stores the resultant power
    temp = []
 
    if (n == 0):
        return 1
 
    # Stores the value of x^(n/2)
    temp = power(x, n // 2)
    if (n % 2 == 0):
        return temp * temp
    else:
        return x * temp * temp
 
# Function to count aN-digit numbers
# whose digits are multiples of x
def count_Total_Numbers(n, x):
     
    total, multiples = 0, 0
 
    # Count adigits which
    # are multiples of x
    for i in range(10):
 
        # Check if current number
        # is a mutiple of X
        if (i % x == 0):
 
            # Increase count of multiples
            multiples += 1
 
    # Check if it's a 1 digit number
    if (n == 1):
        return multiples
 
    # Count the total numbers
    total = ((multiples - 1) *
     power(multiples, n - 1))
 
    # Return the total numbers
    return total
 
# Driver Code
if __name__ == '__main__':
     
    # Given N and X
    N = 1
    X = 3
 
    # Function call
    print(count_Total_Numbers(N, X))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to calculate x^n
// using binary-exponentiation
static int power(int x, int n)
{
     
    // Stores the resultant power
    int temp;
     
    if (n == 0)
        return 1;
     
    // Stores the value of x^(n/2)
    temp = power(x, n / 2);
     
    if (n % 2 == 0)
        return temp * temp;
    else
        return x * temp * temp;
}
 
// Function to count aint N-digit
// numbers whose digits are multiples
// of x
static int count_Total_Numbers(int n,
                               int x)
{
    int total, multiples = 0;
     
    // Count aint digits which
    // are multiples of x
    for(int i = 0; i < 10; i++)
    {
         
        // Check if current number
        // is a mutiple of X
        if (i % x == 0)
     
        // Increase count of multiples
        multiples++;
    }
     
    // Check if it's a 1 digit number
    if (n == 1)
        return multiples;
     
    // Count the total numbers
    total = (multiples - 1) *
             power(multiples, n - 1);
     
    // Return the total numbers
    return total;
}
 
// Driver Code
public static void Main()
{
     
    // Given N and X
    int N = 1, X = 3;
     
    // Function call
    Console.Write(count_Total_Numbers(N, X));
}
}
 
// This code is contributed by sanjoy_62

Javascript

// Javascript program for the above approach
 
// Function to calculate x^n
// using binary-exponentiation
function power(x, n)
{
 
    // Stores the resultant power
    let temp;
 
    if (n == 0)
        return 1;
 
    // Stores the value of x^(n/2)
    temp = power(x, parseInt(n / 2));
    if (n % 2 == 0)
        return temp * temp;
    else
        return x * temp * temp;
}
 
// Function to count all N-digit numbers
// whose digits are multiples of x
function count_Total_Numbers(n, x)
{
    let total, multiples = 0;
 
    // Count all digits which
    // are multiples of x
    for (let i = 0; i < 10; i++)
    {
 
        // Check if current number
        // is a mutiple of X
        if (i % x == 0)
 
            // Increase count of multiples
            multiples++;
    }
 
    // Check if it's a 1 digit number
    if (n == 1)
        return multiples;
 
    // Count the total numbers
    total = (multiples - 1)
            * power(multiples, n - 1);
 
    // Return the total numbers
    return total;
}
 
// Driver Code
 
    // Given N and X
    let N = 1, X = 3;
 
    // Function Call
    document.write(
           count_Total_Numbers(N, X));
 
// This code is contributed by subhammahato348.

输出：

4

时间复杂度： O(Log N)
辅助空间： O(1)