从它们的总和与XOR中找出两个数字

给定两个数X和Y的总和与异或，st和和异或$\in [0, 2^{64}-1]$ ，我们需要找到使X值最小的数字。

例子 ：

Input : S = 17
        X = 13
Output : a = 2
         b = 15

Input : S = 1870807699 
        X = 259801747
Output : a = 805502976
         b = 1065304723

Input : S = 1639
        X = 1176
Output : No such numbers exist

Variables Used:
X ==> XOR of two numbers
S ==> Sum of two numbers
X[i] ==> Value of i-th bit in X
S[i] ==> Value of i-th bit in S

一个简单的解决方案是使用给定的XOR生成所有可能的对。要生成所有对，我们可以遵循以下规则。

1. 如果X [i]为1，则a [i]和b [i]应该都不同，我们有两种情况。
2. 如果X [i]为0，则a [i]和b [i]应该相同。我们有两种情况。

   因此，我们生成2 ^ n个可能的对，其中n是X中的位数。然后对于每对，我们检查其总和是否为S。

   一个有效的解决方案基于以下事实。

   S = X + 2*A
   where A = a AND b

   We can verify above fact using the sum process. In sum, whenever we see both bits 1 (i.e., AND is 1), we make resultant bit 0 and add 1 as carry, which means every bit in AND is left shifted by 1 OR value of AND is multiplied by 2 and added.

   因此我们可以找到A =(S – X)/ 2。

   一旦找到A，就可以使用以下规则找到“ a”和“ b”的所有位。

   1. 如果X [i] = 0且A [i] = 0，则a [i] = b [i] =0。对于该位，只有一种可能性。
   2. 如果X [i] = 0且A [i] = 1，则a [i] = b [i] = 1。
   3. 如果X [i] = 1且A [i] = 0，则(a [i] = 1且b [i] = 0)或(a [i] = 0且b [i] = 1)，我们可以选择两者中的任何一个。
   4. 如果X [i] = 1且A [i] = 1，则结果不可能(注X [i] = 1表示不同的位)

   令总和为S，XOR为X。

   下面是上述方法的实现：

   C++

   // CPP program to find two numbers with
   // given Sum and XOR such that value of
   // first number is minimum.
   #include 
   using namespace std;
     
   // Function that takes in the sum and XOR
   // of two numbers and generates the two 
   // numbers such that the value of X is
   // minimized
   void compute(unsigned long int S, 
               unsigned long int X)
   {
       unsigned long int A = (S - X)/2;
     
       int a = 0, b = 0;
     
       // Traverse through all bits
       for (int i=0; i<8*sizeof(S); i++)
       {
           unsigned long int Xi = (X & (1 << i));
           unsigned long int Ai = (A & (1 << i));
           if (Xi == 0 && Ai == 0)
           {
               // Let us leave bits as 0.
           }
           else if (Xi == 0 && Ai > 0)
           {
               a = ((1 << i) | a); 
               b = ((1 << i) | b); 
           }
           else if (Xi > 0 && Ai == 0)
           {
               a = ((1 << i) | a); 
     
               // We leave i-th bit of b as 0.
           }
           else // (Xi == 1 && Ai == 1)
           {
               cout << "Not Possible";
               return;
           }
       }
     
       cout << "a = " << a << endl << "b = " << b;
   }
     
   // Driver function
   int main()
   {
       unsigned long int S = 17, X = 13;
       compute(S, X);
       return 0;
   }

   Java

   // Java program to find two numbers with 
   // given Sum and XOR such that value of 
   // first number is minimum. 
   class GFG {
     
   // Function that takes in the sum and XOR 
   // of two numbers and generates the two 
   // numbers such that the value of X is 
   // minimized 
   static void compute(long S, long  X) 
   { 
       long A = (S - X)/2; 
       int a = 0, b = 0; 
           final int LONG_FIELD_SIZE     = 8;
     
       // Traverse through all bits 
       for (int i=0; i<8*LONG_FIELD_SIZE; i++) 
       { 
           long Xi = (X & (1 << i)); 
           long Ai = (A & (1 << i)); 
           if (Xi == 0 && Ai == 0) 
           { 
               // Let us leave bits as 0. 
           } 
           else if (Xi == 0 && Ai > 0) 
           { 
               a = ((1 << i) | a); 
               b = ((1 << i) | b); 
           } 
           else if (Xi > 0 && Ai == 0) 
           { 
               a = ((1 << i) | a); 
     
               // We leave i-th bit of b as 0. 
           } 
           else // (Xi == 1 && Ai == 1) 
           { 
               System.out.println("Not Possible"); 
               return; 
           } 
       } 
     
       System.out.println("a = " + a +"\nb = " + b); 
   } 
     
   // Driver function 
       public static void main(String[] args) {
           long S = 17, X = 13; 
       compute(S, X); 
     
       }
   }
   // This code is contributed by RAJPUT-JI

   Python3

   # Python program to find two numbers with
   # given Sum and XOR such that value of
   # first number is minimum.
     
     
   # Function that takes in the sum and XOR
   # of two numbers and generates the two 
   # numbers such that the value of X is
   # minimized
   def compute(S, X):
       A = (S - X)//2
       a = 0
       b = 0
     
       # Traverse through all bits
       for i in range(64):
           Xi = (X & (1 << i))
           Ai = (A & (1 << i))
           if (Xi == 0 and Ai == 0):
               # Let us leave bits as 0.
               pass
                 
           elif (Xi == 0 and Ai > 0):
               a = ((1 << i) | a) 
               b = ((1 << i) | b) 
             
           elif (Xi > 0 and Ai == 0):
               a = ((1 << i) | a) 
               # We leave i-th bit of b as 0.
     
           else: # (Xi == 1 and Ai == 1)
               print("Not Possible")
               return
     
       print("a = ",a)
       print("b =", b)
     
     
   # Driver function
   S = 17
   X = 13
   compute(S, X)
     
   # This code is contributed by ankush_953

   C#

   // C# program to find two numbers with 
   // given Sum and XOR such that value of 
   // first number is minimum. 
   using System;
     
   public class GFG {
     
   // Function that takes in the sum and XOR 
   // of two numbers and generates the two 
   // numbers such that the value of X is 
   // minimized 
   static void compute(long S, long  X) 
   { 
       long A = (S - X)/2; 
       int a = 0, b = 0; 
      
     
       // Traverse through all bits 
       for (int i=0; i<8*sizeof(long); i++) 
       { 
           long Xi = (X & (1 << i)); 
           long Ai = (A & (1 << i)); 
           if (Xi == 0 && Ai == 0) 
           { 
               // Let us leave bits as 0. 
           } 
           else if (Xi == 0 && Ai > 0) 
           { 
               a = ((1 << i) | a); 
               b = ((1 << i) | b); 
           } 
           else if (Xi > 0 && Ai == 0) 
           { 
               a = ((1 << i) | a); 
     
               // We leave i-th bit of b as 0. 
           } 
           else // (Xi == 1 && Ai == 1) 
           { 
               Console.WriteLine("Not Possible"); 
               return; 
           } 
       } 
     
       Console.WriteLine("a = " + a +"\nb = " + b); 
   } 
     
   // Driver function 
       public static void Main() {
           long S = 17, X = 13; 
           compute(S, X); 
       }
   }
   // This code is contributed by RAJPUT-JI

   输出

   a = 15
   b = 2
   

   上述方法的时间复杂度$O(b)$其中b是S中的位数。

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