位数为偶数的数字

如果正整数的数字总和为偶数，则认为该整数是一个好数字。查找第n个最小的有效数。
例子 ：

Input :  n = 1
Output : 2
First good number is smallest positive
number with sum of digits even which is 2.

Input : n = 10
Output : 20

一个简单的解决方案是从1开始遍历所有自然数。对于每个数字x，请检查数字的总和是否为偶数。如果偶数增加好数。最后返回第n个好数字。
一个有效的解决方案基于答案中的模式。让我们列出前20个好数字。前20个有效数字为：2、4、6、8、11、13、15、17、19、20、22、24、26、28、31、33、35、37、39、40。 n的最后一位数字是0到4，答案是2 * n；如果n的最后一位数字是5到9，答案是2 * n + 1。

C++

// C++ program to find n-th
// Good number.
#include 
using namespace std;
 
// Function to find kth good number.
long long int findKthGoodNo(long long int n)
{
    // Find the last digit of n.
    int lastDig = n % 10;
 
    // If last digit is between
    // 0 to 4 then return 2 * n.
    if (lastDig >= 0 && lastDig <= 4)
        return n << 1;
 
    // If last digit is between
    // 5 to 9 then return 2*n + 1.
    else
        return (n << 1) + 1;
}
 
// Driver code
int main()
{
    long long int n = 10;
    cout << findKthGoodNo(n);
    return 0;
}

Java

// Java program to find n-th
// Good number.
class GFG
{
     
    // Function to find kth good number.
    static int findKthGoodNo(int n)
    {
         
        // Find the last digit of n.
        int lastDig = n % 10;
 
        // If last digit is between
        // 0 to 4 then return 2*n.
        if (lastDig >= 0 && lastDig <= 4)
            return n << 1;
 
        // If last digit is between
        // 5 to 9 then return 2*n + 1.
        else
            return (n << 1) + 1;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 10;
         
        System.out.println(findKthGoodNo(n));
    }
}
 
// This code is contributed by
// Smitha Dinesh Semwal

Python 3

# Python 3 program to find
# n-th Good number.
 
# Function to find kth
# good number.
def findKthGoodNo(n):
 
    # Find the last digit of n.
    lastDig = n % 10
 
    # If last digit is between
    # 0 to 4 then return 2 * n.
    if (lastDig >= 0 and lastDig <= 4) :
        return n << 1
 
    # If last digit is between
    # 5 to 9 then return 2 * n + 1.
    else:
        return (n << 1) + 1
 
 
# Driver code
n = 10
print(findKthGoodNo(n))
 
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program to find n-th
// Good number.
using System;
 
class GFG
{
     
    // Function to find kth
    // good number
    public static int findKthGoodNo(int n)
    {
         
        // Find the last digit of n.
        int lastDig = n % 10;
 
        // If last digit is between
        // 0 to 4 then return 2*n.
        if (lastDig >= 0 && lastDig <= 4)
            return n << 1;
 
        // If last digit is between
        // 5 to 9 then return 2*n + 1.
        else
            return (n << 1) + 1;
    }
     
    // Driver code
    static public void Main (string []args)
    {
        int n = 10;
        Console.WriteLine(findKthGoodNo(n));
    }
}
 
// This code is contributed by Ajit.

PHP

= 0 && $lastDig <= 4)
        return $n << 1;
 
    // If last digit is between
    // 5 to 9 then return 2*n + 1.
    else
        return ($n << 1) + 1;
}
 
// Driver code
$n = 10;
echo(findKthGoodNo($n));
 
// This code is contributed by Ajit.
?>

Javascript

 
// JavaScript program to find n-th
// Good number.
 
    // Function to find kth good number.
    function findKthGoodNo(n)
    {
           
        // Find the last digit of n.
        let lastDig = n % 10;
   
        // If last digit is between
        // 0 to 4 then return 2*n.
        if (lastDig >= 0 && lastDig <= 4)
            return n << 1;
   
        // If last digit is between
        // 5 to 9 then return 2*n + 1.
        else
            return (n << 1) + 1;
    }
 
// Driver code
        let n = 10;
        document.write(findKthGoodNo(n));
 
// This code is contributed by souravghosh0416.

输出：

20

时间复杂度： O(1)
辅助空间： O(1)