给定数字N和数字K (1
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：
一种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(例如A)。由于A和B的总和必须为N，因此对应的B为(N – A)。现在检查A和B是否都包含K，如果不包含K，则打印A和B。对于较大的输入(如N = 10 18)，此方法将失败。
时间复杂度： O(N)
辅助空间： O(1)
高效方法：
给定的问题可以更有效地解决。想法是将N的每个数字D分为两个部分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
在这里，A将为2267，而B将为2200。
此方法的详细步骤如下：
1.遍历N并考虑其每个数字D。
2.如果数字D与K相同，则将其分为D1 = D / 2和D2 = D / 2 + D％2。
D％2被添加到D2以确保偶数和奇数D的D1 + D2 =D。
3.否则，将其分为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)之后，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 sapnasingh4991Python3
# 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 divyeshrabadiya07C#
// 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输出：
11671, 22002
时间复杂度： O(M)
辅助空间： O(M)
在此，M是N中的位数。