用0作为数字计数’d’位正整数

给定一个数字d ，它代表一个正整数的位数。求出其中至少有一个零的正整数(精确地由d位数字组成)的总计数。
例子：

Input : d = 1
Output : 0
There's no natural number of 1 digit that
contains a zero.

Input : d = 2
Output : 9
The numbers are, 10, 20, 30, 40, 50, 60, 
                 70, 80 and 90.

我们强烈建议您单击此处并进行实践，然后再继续解决方案。

一种简单的解决方案是遍历所有d位正数。对于每个数字，请遍历其数字，如果有0位数，则递增计数(与此类似)。
以下是一些观察结果：

1. 恰好有d位数字。
2. 最高有效位的数字不能为零(不允许前导零)。
3. 除最高有效位以外的所有其他位都可以包含零。

因此，考虑以上几点，让我们找到具有d位数字的总数：

We can place any of {1, 2, ... 9} in D1
Hence D1 can be filled in 9 ways.

Apart from D1 all the other places can be  10 ways. 
(we can place 0 as well)
Hence the total numbers having d digits can be given as: 
Total =  9*10d-1

Now, let's find the numbers having d digits, that
don't contain zero at any place. 
In this case, all the places can be filled in 9 ways.
Hence count of such numbers is given by:
Non_Zero = 9d

Now the count of numbers having at least one zero 
can be obtained by subtracting Non_Zero from Total.
Hence Answer would be given by:
9*(10d-1 - 9d-1 ) 

以下是相同的程序。

C++

//C++ program to find the count of positive integer of a
// given number of digits that contain atleast one zero
#include
using namespace std;
 
// Returns count of 'd' digit integers have 0 as a digit
int findCount(int d)
{
    return 9*(pow(10,d-1) - pow(9,d-1));
}
 
// Driver Code
int main()
{
    int d = 1;
    cout << findCount(d) << endl;
 
    d = 2;
    cout << findCount(d) << endl;
 
    d = 4;
    cout << findCount(d) << endl;
    return 0;
}

Java

// Java program to find the count
// of positive integer of a
// given number of digits
// that contain atleast one zero
import java.io.*;
 
class GFG {
     
    // Returns count of 'd' digit
    // integers have 0 as a digit
    static int findCount(int d)
    {
        return 9 * ((int)(Math.pow(10, d - 1))
                 - (int)(Math.pow(9, d - 1)));
    }
     
    // Driver Code
    public static void main(String args[])
    {
        int d = 1;
        System.out.println(findCount(d));
         
        d = 2;
        System.out.println(findCount(d));
         
        d = 4;
        System.out.println(findCount(d));
         
    }
}
 
// This code is contributed by Nikita Tiwari.

Python3

# Python 3 program to find the
# count of positive integer of a
# given number of digits that
# contain atleast one zero
import math
 
# Returns count of 'd' digit
# integers have 0 as a digit
def findCount(d) :
    return 9*((int)(math.pow(10,d-1)) - (int)(math.pow(9,d-1)));
 
 
# Driver Code
d = 1
print(findCount(d))
     
d = 2
print(findCount(d))
 
d = 4
print(findCount(d))
 
 
# This code is contributed by Nikita Tiwari.

C#

// C# program to find the count
// of positive integer of a
// given number of digits
// that contain atleast one zero.
using System;
 
class GFG {
     
    // Returns count of 'd' digit
    // integers have 0 as a digit
    static int findCount(int d)
    {
        return 9 * ((int)(Math.Pow(10, d - 1))
                 - (int)(Math.Pow(9, d - 1)));
    }
     
    // Driver Code
    public static void Main()
    {
        int d = 1;
        Console.WriteLine(findCount(d));
         
        d = 2;
        Console.WriteLine(findCount(d));
         
        d = 4;
        Console.WriteLine(findCount(d));
         
    }
}
 
// This code is contributed by nitin mittal.

Javascript

 
// JavaScript program to find the count
// of positive integer of a
// given number of digits
// that contain atleast one zero
 
    // Returns count of 'd' digit
    // integers have 0 as a digit
    function findCount(d)
    {
        return 9 * ((Math.pow(10, d - 1))
                 - (Math.pow(9, d - 1)));
    }
 
// Driver Code
 
        let d = 1;
        document.write(findCount(d) + "<br/>");
           
        d = 2;
        document.write(findCount(d) + "<br/>");
           
        d = 4;
        document.write(findCount(d) + "<br/>");   
     
    // This code is contributed by target_2.

输出 ：

0
9
2439