找出总和为 N 的两个数字，使得它们都不包含数字 K

给定一个数字N和一个数字K (1 < K <= 9)，任务是找到两个整数A和B ，使得A + B = N并且数字 K不存在于 A 或 B 中。

例子：

Input: N = 443, K = 4 
Output: 256, 187 
Explanation: 
Sum = 256 + 187 = 443 = N 
Also, K(4) is not present in 256 or 187 
Hence, 256 and 187 are valid output.

Input: N = 678, K = 9 
Output: 311, 367 
Explanation: 
Sum = 311 + 367 = 678 = N 
Also, K(9) is not present in 311 or 367 
Hence, 311 and 367 are valid output. 

天真的方法：
一个简单的方法是运行一个循环并考虑从 1 到 N-1 的每个元素（比如 A）。由于 A 和 B 的总和必须是 N，因此对应的 B 将是 (N – A)。现在检查 A 和 B 是否都包含 K，如果不包含，则打印 A 和 B。对于较大的输入，如 N = 10 ^{18 ，}这种方法将失败。

时间复杂度： O(N)
辅助空间： O(1)

有效的方法：
可以更有效地解决给定的问题。这个想法是将 N 的每个数字D分成两部分D1和D2 ，使得 D1 + D2 = D。取两个字符串A和B ，A 将包含所有 D1，而字符串B 将包含所有 D2。
例如，N = 4467 和 K = 4

D    D1    D2
4    2    2
4    2    2
6    6    0
7    7    0    

这里，A 将是 2267，B 将是 2200。
该方法的详细步骤如下：
1.遍历N并考虑它的每个数字D 。
2.如果数字 D 与 K 相同，则将其分解为 D1 = D / 2 和 D2 = D / 2 + D % 2。
D % 2 被添加到 D2 以确保 D1 + D2 = D 对于偶数和奇数 D。
3.否则，将其分解为 D1 = D 和 D2 = 0。
4.继续将 D1 附加到字符串A 并将 D2 附加到字符串B。
5.最后，在删除前导零（如果有）后打印字符串A 和 B。

下面是上述方法的实现：

C++

// C++ implementation of
// the above approach
 
#include 
using namespace std;
 
string removeLeadingZeros(string str)
{
    // Count leading zeros
    int i = 0;
    int n = str.length();
    while (str[i] == '0' && i < n)
        i++;
 
    // It removes i characters
    // starting from index 0
    str.erase(0, i);
 
    return str;
}
 
void findPairs(int sum, int K)
{
 
    string A, B;
    A = "";
    B = "";
 
    string N = to_string(sum);
    int n = N.length();
 
    // Check each digit of the N
    for (int i = 0; i < n; i++) {
 
        int D = N[i] - '0';
 
        // If digit is K break it
        if (D == K) {
            int D1, D2;
            D1 = D / 2;
 
            // For odd numbers
            D2 = D / 2 + D % 2;
 
            // Add D1 to A and D2 to B
            A = A + char(D1 + '0');
            B = B + char(D2 + '0');
        }
 
        // If the digit is not K,
        // no need to break
        // string D in A and 0 in B
        else {
            A = A + char(D + '0');
            B = B + '0';
        }
    }
 
    // Remove leading zeros
    A = removeLeadingZeros(A);
    B = removeLeadingZeros(B);
 
    // Print the answer
    cout << A << ", " << B << endl;
}
 
// Driver code
int main()
 
{
    int N = 33673;
    int K = 3;
 
    findPairs(N, K);
 
    return 0;
}

Java

// Java implementation of the above approach
class GFG{
 
static String removeLeadingZeros(String str)
{
     
    // Count leading zeros
    int i = 0;
    int n = str.length();
     
    while (str.charAt(i) == '0' && i < n)
        i++;
 
    // It removes i characters
    // starting from index 0
    str = str.substring(i);
 
    return str;
}
 
static void findPairs(int sum, int K)
{
 
    String A, B;
    A = "";
    B = "";
 
    String N = String.valueOf(sum);
    int n = N.length();
 
    // Check each digit of the N
    for(int i = 0; i < n; i++)
    {
       int D = N.charAt(i) - '0';
        
       // If digit is K break it
       if (D == K)
       {
           int D1, D2;
           D1 = D / 2;
            
           // For odd numbers
           D2 = D / 2 + D % 2;
            
           // Add D1 to A and D2 to B
           A = A + (char)(D1 + '0');
           B = B + (char)(D2 + '0');
       }
        
       // If the digit is not K,
       // no need to break
       // String D in A and 0 in B
       else
       {
           A = A + (char)(D + '0');
           B = B + '0';
       }
    }
     
    // Remove leading zeros
    A = removeLeadingZeros(A);
    B = removeLeadingZeros(B);
 
    // Print the answer
    System.out.print(A + ", " + B + "\n");
}
 
// Driver code
public static void main(String[] args)
{
    int N = 33673;
    int K = 3;
 
    findPairs(N, K);
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 implementation of the
# above approach
def removeLeadingZeros(Str):
     
    # Count leading zeros
    i = 0
    n = len(Str)
       
    while (Str[i] == '0' and i < n):
        i += 1
   
    # It removes i characters
    # starting from index 0
    Str = Str[i:]
    return Str
   
def findPairs(Sum, K):
  
    A = ""
    B = ""
   
    N = str(Sum)
    n = len(N)
 
    # Check each digit of the N
    for i in range(n):
     
       D = int(N[i]) - int('0')
          
       # If digit is K break it
       if (D == K):
           D1 = D // 2;
              
           # For odd numbers
           D2 = (D // 2) + (D % 2)
 
           # Add D1 to A and D2 to B
           A = A + (str)(D1 + int('0'))
           B = B + (str)(D2 + int('0'))
            
       # If the digit is not K,
       # no need to break
       # String D in A and 0 in B
       else:
           A = A + (str)(D + int('0'))
           B = B + '0'
     
    # Remove leading zeros
    A = removeLeadingZeros(A)
    B = removeLeadingZeros(B)
   
    # Print the answer
    print(A + ", " + B)
   
# Driver code
N = 33673
K = 3
 
findPairs(N, K)
 
# This code is contributed divyeshrabadiya07

C#

// C# implementation of the above approach
using System;
 
class GFG{
  
static String removeLeadingZeros(String str)
{
      
    // Count leading zeros
    int i = 0;
    int n = str.Length;
      
    while (str[i] == '0' && i < n)
        i++;
  
    // It removes i characters
    // starting from index 0
    str = str.Substring(i);
  
    return str;
}
  
static void findPairs(int sum, int K)
{
  
    String A, B;
    A = "";
    B = "";
  
    String N = String.Join("", sum);
    int n = N.Length;
  
    // Check each digit of the N
    for(int i = 0; i < n; i++)
    {
       int D = N[i] - '0';
         
       // If digit is K break it
       if (D == K)
       {
           int D1, D2;
           D1 = D / 2;
             
           // For odd numbers
           D2 = D / 2 + D % 2;
             
           // Add D1 to A and D2 to B
           A = A + (char)(D1 + '0');
           B = B + (char)(D2 + '0');
       }
         
       // If the digit is not K,
       // no need to break
       // String D in A and 0 in B
       else
       {
           A = A + (char)(D + '0');
           B = B + '0';
       }
    }
      
    // Remove leading zeros
    A = removeLeadingZeros(A);
    B = removeLeadingZeros(B);
  
    // Print the answer
    Console.Write(A + ", " + B + "\n");
}
  
// Driver code
public static void Main(String[] args)
{
    int N = 33673;
    int K = 3;
  
    findPairs(N, K);
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
// JavaScript implementation of
// the above approach
 
function removeLeadingZeros(str)
{
    // Count leading zeros
    var i = 0;
    var n = str.length;
    while (str[i] == '0' && i < n)
        i++;
 
    return str;
}
 
function findPairs(sum, K)
{
 
    var A, B;
    A = "";
    B = "";
 
    var N = (sum.toString());
    var n = N.length;
 
    // Check each digit of the N
    for (var i = 0; i < n; i++) {
 
        var D = N[i] - '0';
 
        // If digit is K break it
        if (D == K) {
            var D1, D2;
            D1 = parseInt(D / 2);
 
            // For odd numbers
            D2 = parseInt(D / 2) + D % 2;
 
            // Add D1 to A and D2 to B
            A = A +
            String.fromCharCode(D1 + '0'.charCodeAt(0));
             
            B = B +
            String.fromCharCode(D2 + '0'.charCodeAt(0));
        }
 
        // If the digit is not K,
        // no need to break
        // string D in A and 0 in B
        else {
            A = A +
            String.fromCharCode(D + '0'.charCodeAt(0));
            B = B + '0';
        }
    }
 
    // Remove leading zeros
    A = removeLeadingZeros(A);
    B = removeLeadingZeros(B);
 
    // Print the answer
    document.write( A + ", " + B );
}
 
// Driver code
var N = 33673;
var K = 3;
findPairs(N, K);
 

输出：

11671, 22002

时间复杂度： O(M)
辅助空间： O(M)
其中，M 是 N 中的位数。