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We are given two numbers S and K. The goal is to find ordered pairs of positive numbers such that their sum is S and XOR is K.

We will do this by starting from i=1 to i<S-1 and j=i+1 to j<S. If any pair (i,j) has sum==S and i^j==K (XOR). Then increment count by 2 as (i,j) and (j,i) both will be counted as different pairs.

Let’s understand with examples.

**Input**

S=10 K=4

**Output**

Ordered pairs such that sum is S and XOR is K: 2

**Explanation**

Pairs will be (3,7) and (7,3)

**Input**

S=12 K=6

**Output**

Ordered pairs such that sum is S and XOR is K: 0

**Explanation**

No such pairs possible.

We take integers S and K.

Function sumXOR(int s, int k) takes s and k and returns the count of ordered pairs with sum=s and xor=k

Take the initial variable count as 0 for pairs.

Traverse using two for loops for making pairs.

Start from i=1 to i<s-1. And j=i+1 to j<s-1.

Now for each pair (i,j) check if (i+j==s) && (i^j==k). If true increment count by 2 as (i,j) and (j,i) both are different pairs.

At the end of all loops count will have a total number of such pairs.

Return the count as result.

#include <bits/stdc++.h> using namespace std; int sumXOR(int s, int k){ int count = 0; for (int i = 1; i < s; i++){ for(int j=i+1; j<s-1; j++){ if( (i+j)==s && (i^j)==k){ count+=2; //(i,j) and (j,i) are two pairs } } } return count; } int main(){ int S = 9, K = 5; cout <<"Ordered pairs such that sum is S and XOR is K: "<< sumXOR(S,K); return 0; }

If we run the above code it will generate the following output −

Ordered pairs such that sum is S and XOR is K: 4

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