查找小于或等于N且具有奇数位数的正整数的数量

给定一个整数N(其中1≤N≤10 ⁵⁾ ，任务是找到小于或等于N且具有奇数位数且不带前导零的正整数。

例子：

Input: N = 11
Output: 9
1, 2, 3, …, 8 and 9 are the numbers ≤ 11
with odd number of digits.

Input: N = 893
Output: 803

天真的方法：从1遍历到N，并检查每个数字是否包含奇数位。

高效方法：对于价值观：

• 当N <10时，有效数字的计数将为N。
• 当N / 10 <10时为9 。
• 如果N / 100 <10，则9 + N – 99 。
• 当N / 1000 <10时为9 + 900 。
• 当N / 10000 <10时，则为909 + N – 9999 。
• 否则为90909 。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the number of
// positive integers less than or equal
// to N that have odd number of digits
int odd_digits(int n)
{
    if (n < 10)
        return n;
    else if (n / 10 < 10)
        return 9;
    else if (n / 100 < 10)
        return 9 + n - 99;
    else if (n / 1000 < 10)
        return 9 + 900;
    else if (n / 10000 < 10)
        return 909 + n - 9999;
    else
        return 90909;
}
  
// Driver code
int main()
{
    int n = 893;
  
    cout << odd_digits(n);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG 
{
  
// Function to return the number of
// positive integers less than or equal
// to N that have odd number of digits
static int odd_digits(int n)
{
    if (n < 10)
        return n;
    else if (n / 10 < 10)
        return 9;
    else if (n / 100 < 10)
        return 9 + n - 99;
    else if (n / 1000 < 10)
        return 9 + 900;
    else if (n / 10000 < 10)
        return 909 + n - 9999;
    else
        return 90909;
}
  
// Driver code
public static void main(String []args) 
{
    int n = 893;
  
    System.out.println(odd_digits(n));
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach 
  
# Function to return the number of 
# positive integers less than or equal 
# to N that have odd number of digits 
def odd_digits(n) :
  
    if (n < 10) :
        return n; 
    elif (n / 10 < 10) :
        return 9; 
    elif (n / 100 < 10) :
        return 9 + n - 99; 
    elif (n / 1000 < 10) :
        return 9 + 900; 
    elif (n / 10000 < 10) : 
        return 909 + n - 9999; 
    else :
        return 90909; 
  
# Driver code 
if __name__ == "__main__" : 
  
    n = 893; 
  
    print(odd_digits(n)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
                      
class GFG 
{
  
// Function to return the number of
// positive integers less than or equal
// to N that have odd number of digits
static int odd_digits(int n)
{
    if (n < 10)
        return n;
    else if (n / 10 < 10)
        return 9;
    else if (n / 100 < 10)
        return 9 + n - 99;
    else if (n / 1000 < 10)
        return 9 + 900;
    else if (n / 10000 < 10)
        return 909 + n - 9999;
    else
        return 90909;
}
  
// Driver code
public static void Main(String []args) 
{
    int n = 893;
  
    Console.WriteLine(odd_digits(n));
}
}
  
// This code is contributed by 29AjayKumar

输出：

803