用N位数字计数数字，该数字由0的奇数组成

给我们一个数字N。任务是找到具有N个数字和零的奇数的数字的计数。
注意：数字可以以0开头。

例子：

Input : N = 2
Output : Count = 18

Input : N = 3
Output : Count = 244

假设一个N位数字仅包含一个零。因此，只能以1种方式填充数字中的数字为零，其余位置可以用9种不同的方式填充从1到9的数字。因此，用N位数字对所有此类数字进行计数，并且只有1个零＝ ^N C ₁ *(9 ^N-1 ) 。
类似地，所有具有N位数字和3个零的数字的计数= ^N C ₃ *(9 ^N-3 ) 。
等等。
因此，所有具有N位数字和奇数个零的数字的计数将为，

^NC₁*(9^N-1) + ^NC₃*(9^N-3) + ^NC₅*(9^N-5) +…….+ ^NC_K*(9^N-K)
Where, K is an odd number less than N. 

上面的等式可以写成：

(9^N)((^NC₁ * (1/9)) + (^NC₃ * (1/9^3)) + (^NC₅ * (1/9^5)) +… 
 

上式可以表示为两个序列的减法，(9 ^N )* {((1 + x) ^N- (1-x) ^N } / 2，其中x = 1/9
等于

(10N - 8N)/2

下面是上述方法的实现：

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) - pow(8, N)) / 2;
}
 
// Driver code
int main()
{
    int n = 5;
 
    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.io.*;
 
class GFG {
 
    // Function to count Numbers
    // with N digits which consists
    // of odd number of 0's
    static int countNumbers(int N)
    {
        return (int)(Math.pow(10, N)
                     - Math.pow(8, N)) / 2;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 5;
        System.out.println(countNumbers(n));
    }
}
 
// This code is contributed by Shashank

Python 3

# Python 3 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 countNumbers(N):
 
    return (pow(10, N) - pow(8, N)) // 2
 
 
# Driver code
if __name__ == "__main__":
 
    n = 5
 
    print(countNumbers(n))
 
# This code is contributed
# by ChitraNayal

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 int countNumbers(int N)
    {
        return (int)(Math.Pow(10, N)
                     - Math.Pow(8, N)) / 2;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 5;
        Console.WriteLine(countNumbers(n));
    }
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

Javascript

    // Javascript 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
    function countNumbers(N)
    {
        return parseInt((Math.pow(10, N) - Math.pow(8, N)) / 2, 10);
    }
     
    let n = 5;
    document.write(countNumbers(n));
     
    // This code is contributed by divyeshrabadiya07.
     

输出

33616

注意：答案可能非常大，因此对于大于9的N，请使用模幂。

时间复杂度： O(log n)
辅助空间： O(1)

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