算法设计：计算d位数的正整数，以0作为数字

本文概述

• C ++
• Java
• Python3
• C#
• 的PHP

给定一个数字d, 代表正整数的位数。求出其中至少有一个零的正整数(精确地由d个数字组成)的总数。

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.

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；如果有任何0数字, 则递增计数(类似于这个)。

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置可以包含零。

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<bits/stdc++.h>
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.

的PHP

<?php
//PHP 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 * (pow(10, $d - 1) - 
                 pow(9, $d - 1));
}
  
//Driver Code
{
     $d = 1;
     echo findCount( $d ), "\n" ;
  
     $d = 2;
     echo findCount( $d ), "\n" ;
  
     $d = 4;
     echo findCount( $d ), "\n" ;
     return 0;
}
  
//This code is contributed by nitin mittal
?>

输出：

0
9
2439

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