最小化 A 和 B 的数字之和，使得 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并且A和B的数字之和最小。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的数字之和。很明显，答案不能小于N的数字之和，因为每当产生进位时数字之和就会减少。而且，当N是10的幂时，显然答案不可能是1 ，所以答案是10 。因为A或B不能为0，因为它们都必须是正数。
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

Javascript

 
// Javascript implementation of the approach
 
// Function to return the minimum
// possible sum of digits of A
// and B such that A + B = n
function minSum(n)
{
 
    // Find the sum of digits of n
    var sum = 0;
    while (n > 0) {
        sum += (n % 10);
        n = parseInt(n/10);
    }
 
    // If num is a power of 10
    if (sum == 1)
        return 10;
 
    return sum;
}
 
// Driver code
var n = 1884;
document.write( minSum(n));
 
// This code is contributed by famously.

输出：

21

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