# 算法设计：计算d位数的正整数，以0作为数字

2021年4月10日12:56:27 发表评论 645 次浏览

## 本文概述

``````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.``````

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。

``````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 )``````

## 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
?>``````

``````0
9
2439`````` 