将 N 拆分为两个整数，将 A 和 B 相加使它们相等

给定三个正整数A 、 B和N ，任务是将N分成两部分，使它们相等，即找到两个正整数X和Y ，使得X + Y = N和A + X = B + Y 。如果不存在这样的对，则打印-1 。

例子：

Input: A = 1, B = 3, N = 4 
Output: 3 1 
Explanation: If X = 3 and Y = 1, then A + X = B + Y and X + Y =4

Input: A = 1, B = 3, N = 1 
Output: -1 
 

天真的方法：
解决这个问题的最简单的方法是生成所有可能的总和为N 的对，并检查每对，如果A + X = B + Y 。

时间复杂度： O(N ² )
辅助空间： O(1)

有效的方法：
可以观察到，由于X + Y = N和A + X = B + Y ，那么 X 可以表示为(N + B – A) / 2 。只需检查(N + B – A) / 2是否为偶数。如果是偶数，计算X和对应的Y 。否则，打印-1 。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to calculate the
// splitted numbers
void findPair(int A, int B, int N)
{
    int X, Y;
 
    // Calculate X
    X = N - B + A;
 
    // If X is odd
    if (X % 2 != 0) {
 
        // No pair is possible
        cout << "-1";
    }
 
    // Otherwise
    else {
 
        // Calculate X
        X = X / 2;
 
        // Calculate Y
        Y = N - X;
        cout << X << " " << Y;
    }
}
 
// Driver Code
int main()
{
    int A = 1;
    int B = 3;
    int N = 4;
    findPair(A, B, N);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
     
// Function to calculate the
// splitted numbers
static void findPair(int A, int B, int N)
{
    int X, Y;
 
    // Calculate X
    X = N - B + A;
 
    // If X is odd
    if (X % 2 != 0)
    {
 
        // No pair is possible
        System.out.print("-1");
    }
 
    // Otherwise
    else
    {
         
        // Calculate X
        X = X / 2;
 
        // Calculate Y
        Y = N - X;
        System.out.print(X + " " + Y);
    }
}
 
//Driver function
public static void main (String[] args)
{
    int A = 1;
    int B = 3;
    int N = 4;
     
    findPair(A, B, N);
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to implement
# the above approach
 
# Function to calculate the
# splitted numbers
def findPair(A, B, N):
 
    # Calculate X
    X = N - B + A
 
    # If X is odd
    if (X % 2 != 0):
 
        # No pair is possible
        print("-1")
     
    # Otherwise
    else :
 
        # Calculate X
        X = X // 2
 
        # Calculate Y
        Y = N - X
        print (X , Y)
 
# Driver Code
if __name__ == "__main__":
 
    A = 1
    B = 3
    N = 4
     
    findPair(A, B, N)
 
# This code is contributed by chitranayal

C#

// C# program to implement 
// the above approach 
using System;
   
class GFG{
       
// Function to calculate the 
// splitted numbers 
static void findPair(int A, int B, int N) 
{ 
    int X, Y; 
   
    // Calculate X 
    X = N - B + A; 
   
    // If X is odd 
    if (X % 2 != 0) 
    { 
         
        // No pair is possible 
        Console.Write("-1"); 
    } 
   
    // Otherwise 
    else
    { 
           
        // Calculate X 
        X = X / 2; 
   
        // Calculate Y 
        Y = N - X; 
        Console.Write(X + " " + Y); 
    } 
}
   
// Driver code
public static void Main(string[] args)
{
    int A = 1; 
    int B = 3; 
    int N = 4; 
       
    findPair(A, B, N); 
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// Javascript program to implement
// the above approach
 
    // Function to calculate the
    // splitted numbers
    function findPair(A , B , N) {
        var X, Y;
 
        // Calculate X
        X = N - B + A;
 
        // If X is odd
        if (X % 2 != 0) {
 
            // No pair is possible
            document.write("-1");
        }
 
        // Otherwise
        else {
 
            // Calculate X
            X = X / 2;
 
            // Calculate Y
            Y = N - X;
            document.write(X + " " + Y);
        }
    }
 
    // Driver function
     
        var A = 1;
        var B = 3;
        var N = 4;
 
        findPair(A, B, N);
 
// This code contributed by umadevi9616
 

输出：

1 3

时间复杂度： O(1)
辅助空间： O(1)