不包含9的n位数字的数量

给定数字n，找出可以形成多少个不包含9的n位数字。

例子：

Input : 1
Output : 8
Explanation :
Except 9, all numbers are possible

Input : 2
Output : 72
Explanation :
Except numbers from 90 - 99 and all
two digit numbers that does not end
with 9 are possible.

可以形成的n位数字的总数将为9 * 10 ^(n-1)，除了第一个位置以外，所有数字都可以用10个数字(0-9)填充。如果从列表中删除了数字9，则n位数字的总数将为8 * 9 ^(n-1)。

下面是上述想法的实现：

C++

// CPP program to find number of n 
// digit numbers that do not 
// contain 9 as it's digit
#include 
using namespace std;
  
// function to find number of 
// n digit numbers possible 
int totalNumber(int n)
{
    return 8*pow(9, n - 1);
}
  
// driver function
int main()
{
    int n = 3;
    cout << totalNumber(n);
    return 0;
}

Java

// Java program to find number of  
// n digit numbers that do not 
// contain 9 as it's digit
import java.io.*;
  
public class GFG
{
      
// function to find number of 
// n digit numbers possible 
static int totalNumber(int n)
{
    return 8 * (int)Math.pow(9, n - 1);
}
  
    // Driver Code
    static public void main (String[] args)
    {
        int n = 3;
        System.out.println(totalNumber(n));
          
    }
}
  
// This code is contributed by vt_m.

Python3

# python program to find number of n 
# digit numbers that do not 
# contain 9 as it's digit
  
# function to find number of 
# n digit numbers possible 
def totalNumber(n):
      
    return 8 * pow(9, n - 1);
   
# driver function
n = 3
print(totalNumber(n))
  
# This code is contributed by Sam007

C#

// C# program to find number of  
// n digit numbers that do not 
// contain 9 as it's digit
using System;
  
public class GFG
{
      
// function to find number of 
// n digit numbers possible 
static int totalNumber(int n)
{
    return 8 * (int)Math.Pow(9, n - 1);
}
  
    // Driver Code
    static public void Main ()
    {
        int n = 3;
        Console.WriteLine(totalNumber(n));
          
    }
}
  
// This code is contributed by vt_m.

输出：

648