计数N位数字，其中包括0的偶数

给定数字N。任务是找到具有N位数字和零的偶数的数字的计数。

注意：数字可以以0开头。

例子：

Input: N = 2
Output: Count = 81 
Total 2 digit numbers are 99 considering 1 as 01.
2 digit numbers are 01, 02, 03, 04, 05.... 99
Numbers with odd 0's are 01, 02, 03, 04, 05, 06, 07, 08, 09
10, 20, 30, 40, 50, 70, 80, 90 i.e. 18
The rest of the numbers between 01 and 99 will 
do not have any zeroes and zero is also an even number.
So, numbers with even 0's are 99 - 18 = 81.

Input: N = 3
Output: Count = 755

方法：这个想法是找到由0的奇数组成的N位数的计数，然后从N位数的总数中减去它，得到偶数为0的数字。

C++

// C++ program to count numbers with N digits
// which consists of odd number of 0's
#include 
using namespace std;
  
// Function to count Numbers with N digits
// which consists of odd number of 0's
int countNumbers(int N)
{
    return (pow(10, N) - 1) - (pow(10, N) - pow(8, N)) / 2;
}
  
// Driver code
int main()
{
    int n = 2;
  
    cout << countNumbers(n) << endl;
  
    return 0;
}

Java

// Java program to count numbers 
// with N digits which consists 
// of odd number of 0's 
import java.lang.*;
import java.util.*;
  
class GFG
{
      
// Function to count Numbers with 
// N digits which consists of odd 
// number of 0's 
static double countNumbers(int N) 
{ 
    return (Math.pow(10, N) - 1) - 
           (Math.pow(10, N) - 
            Math.pow(8, N)) / 2; 
} 
  
// Driver code 
static public void main (String args[])
{
    int n = 2; 
    System.out.println(countNumbers(n)); 
}
}
  
// This code si contributed 
// by Akanksha Rai

Python3

# Python3  program to count numbers with N digits
#  which consists of odd number of 0's
  
# Function to count Numbers with N digits
#  which consists of odd number of 0's
def countNumber(n):
  
    return (pow(10,n)-1)- (pow(10,n)-pow(8,n))//2
  
  
# Driver code
n = 2
print(countNumber(n))
  
# This code is contributed by Shrikant13

C#

// C# program to count numbers 
// with N digits which consists 
// of odd number of 0's 
using System;
  
class GFG
{
      
// Function to count Numbers with 
// N digits which consists of odd 
// number of 0's 
static double countNumbers(int N) 
{ 
    return (Math.Pow(10, N) - 1) - 
           (Math.Pow(10, N) - 
            Math.Pow(8, N)) / 2; 
} 
  
// Driver code 
static public void Main ()
{
    int n = 2; 
    Console.WriteLine(countNumbers(n)); 
}
}
  
// This code si contributed by ajit

输出：

81

注意：答案可能非常大，因此对于大于9的N，请使用模幂。