最小化A和B的数字总和，以使A + B = N

给定一个整数N ，任务是找到两个正整数A和B ，使得A + B = N且A和B的数字总和最小。打印A和B的数字总和。

例子：

Input: N = 16
Output: 7
(10 + 6) = 16 and (1 + 0 + 6) = 7
is minimum possible.

Input: N = 1000
Output: 10
(900 + 100) = 1000

方法：如果N是10的幂，那么答案将是10，否则答案将是N的数字总和。显然，答案不能小于N的位数之和，因为每当产生进位时位数的总和就会减少。此外，当N为10的幂时，答案显然不能为1 ，因此答案将为10 。由于A或B不能为0，因为它们两个都必须为正数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the minimum
// possible sum of digits of A
// and B such that A + B = n
int minSum(int n)
{
    // Find the sum of digits of n
    int sum = 0;
    while (n > 0) {
        sum += (n % 10);
        n /= 10;
    }
  
    // If num is a power of 10
    if (sum == 1)
        return 10;
  
    return sum;
}
  
// Driver code
int main()
{
    int n = 1884;
  
    cout << minSum(n);
  
    return 0;
}

Java

// Java implementation of the approach
  
class GFG
{
  
// Function to return the minimum
// possible sum of digits of A
// and B such that A + B = n
static int minSum(int n)
{
    // Find the sum of digits of n
    int sum = 0;
    while (n > 0)
    {
        sum += (n % 10);
        n /= 10;
    }
  
    // If num is a power of 10
    if (sum == 1)
        return 10;
  
    return sum;
}
  
// Driver code
public static void main(String[] args)
{
    int n = 1884;
  
    System.out.print(minSum(n));
  
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python implementation of the approach 
  
# Function to return the minimum 
# possible sum of digits of A 
# and B such that A + B = n 
def minSum(n) : 
  
    # Find the sum of digits of n 
    sum = 0; 
    while (n > 0) :
        sum += (n % 10); 
        n //= 10; 
  
    # If num is a power of 10 
    if (sum == 1) :
        return 10; 
  
    return sum; 
  
# Driver code 
if __name__ == "__main__" : 
    n = 1884; 
  
    print(minSum(n)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to return the minimum
// possible sum of digits of A
// and B such that A + B = n
static int minSum(int n)
{
    // Find the sum of digits of n
    int sum = 0;
    while (n > 0)
    {
        sum += (n % 10);
        n /= 10;
    }
  
    // If num is a power of 10
    if (sum == 1)
        return 10;
  
    return sum;
}
  
// Driver code
public static void Main(String[] args)
{
    int n = 1884;
  
    Console.Write(minSum(n));
}
}
  
// This code is contributed by PrinciRaj1992

输出：

21