求小于等于 N 的正整数中有奇数位数的个数

原文:https://www . geesforgeks . org/find-小于或等于 n 的正整数位数/T1

给定一个整数 N ，其中 1 ≤ N ≤ 10 5 ，任务是找出小于或等于 N 的正整数个数，其位数为奇数，没有前导零。 举例:

输入: N = 11 输出: 9 1、2、3、…、8、9 为奇数位数的数字≤ 11 。 输入: N = 893 输出: 803

简单方法:从 1 到 N 遍历，检查每个数字是否包含奇数。 有效方法:为数值:

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

以下是上述方法的实现:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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 语言(一种计算机语言，尤用于创建网站)

// 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

Python 3

# 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

java 描述语言

<script>
// Java script implementation of the approach

// Function to return the number of
// positive integers less than or equal
// to N that have odd number of digits
function odd_digits( 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
let n = 893;

    document.write(odd_digits(n));

// This code is contributed by sravan kumar Gottumukkala
</script>

Output: 

803

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