将数字分为两部分，以使数字总和最大

给定数字N。任务是找到SumOfDigits(A)+ SumOfDigits(B)的最大可能值，以使A + B = n(0 <= A，B <= n)。

例子：

Input: N = 35
Output: 17
35 = 9 + 26
SumOfDigits(26) = 8, SumOfDigits(9) = 9
So, 17 is the answer.

Input: N = 7
Output: 7

方法：将数字分为A和B两部分，以使A等于9，即最接近N的数字，并且B = NA。例如，N = 35，最接近35的最小数字为29。

C++

// C++ implementation of above approach
#include 
using namespace std;
  
// Returns sum of digits of x
int sumOfDigitsSingle(int x)
{
    int ans = 0;
    while (x) {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}
  
// Returns closest number to x in terms of 9's.
int closest(int x)
{
    int ans = 0;
    while (ans * 10 + 9 <= x)
        ans = ans * 10 + 9;
  
    return ans;
}
  
int sumOfDigitsTwoParts(int N)
{
    int A = closest(N);
    return sumOfDigitsSingle(A) + sumOfDigitsSingle(N - A);
}
  
// Driver code
int main()
{
    int N = 35;
    cout << sumOfDigitsTwoParts(N);
    return 0;
}

Java

// Java implementation of above approach
// Returns sum of digits of x
import java.util.*;
import java.lang.*;
import java.io.*;
  
class GFG
{
static int sumOfDigitsSingle(int x)
{
    int ans = 0;
    while (x != 0)
    {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}
  
// Returns closest number to x 
// in terms of 9's.
static int closest(int x)
{
    int ans = 0;
    while (ans * 10 + 9 <= x)
        ans = ans * 10 + 9;
  
    return ans;
}
  
static int sumOfDigitsTwoParts(int N)
{
    int A = closest(N);
    return sumOfDigitsSingle(A) + 
           sumOfDigitsSingle(N - A);
}
  
// Driver code
public static void main(String args[])
{
    int N = 35;
    System.out.print(sumOfDigitsTwoParts(N));
}
}
  
// This code is contributed by
// Subhadeep Gupta

Python 3

# Python 3 implementation of above approach 
  
#  Returns sum of digits of x 
def sumOfDigitsSingle(x) :
    ans = 0
    while x :
        ans += x % 10
        x //= 10
  
    return ans
  
# Returns closest number to x in terms of 9's
def closest(x) :
    ans = 0
    while (ans * 10 + 9 <= x) :
        ans = ans * 10 + 9
  
    return ans
  
def sumOfDigitsTwoParts(N) :
    A = closest(N)
  
    return sumOfDigitsSingle(A) + sumOfDigitsSingle(N - A)
  
  
# Driver Code
if __name__ == "__main__" :
  
    N = 35
    print(sumOfDigitsTwoParts(N))
  
# This code is contributed by ANKITRAI1

C#

// C# implementation of above approach
// Returns sum of digits of x
  
using System; 
class GFG
{
static int sumOfDigitsSingle(int x)
{
    int ans = 0;
    while (x != 0)
    {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}
   
// Returns closest number to x 
// in terms of 9's.
static int closest(int x)
{
    int ans = 0;
    while (ans * 10 + 9 <= x)
        ans = ans * 10 + 9;
   
    return ans;
}
   
static int sumOfDigitsTwoParts(int N)
{
    int A = closest(N);
    return sumOfDigitsSingle(A) + 
           sumOfDigitsSingle(N - A);
}
   
// Driver code
public static void Main()
{
    int N = 35;
    Console.Write(sumOfDigitsTwoParts(N));
}
}

输出：

17