Find two integers A and B such that A ^ N = A + N and B ^ N = B + N
Given a non-negative integer N, the task is to find two integers A (greatest integer smaller than N) and (smallest integer greater than N) such that A + N = A ^ N and B + N = B ^ N
Examples:
Input: N = 5
Output: A = 2 and B = 8
2 + 8 = 2 ^ 8 = 10Input: N = 10
Output: A = 5 and B = 16
5 + 16 = 5 ^ 16 = 21
Approach: Lets find A and B independently. To solve this problem we have to use the property, x + y = x^y + 2 * (x & y)
Since the problem states that xor sum is equal to the given sum which implies that their AND must be 0.
- Finding A: N can be represented as a series of bits of 0 and 1. To find A we will first have to find the most significant bit of N which is set. Since A & N = 0, The places where N has set bit, for that places we will make bits of A as unset and for the places where N has unset bit, we will make that bit set for A as we want to maximize A. This we will do for all the bits from most significant to the least significant. Hence we will get our A.
- Finding B: Finding B is easy. Let i be the position of the leftmost set bit in 1. We want B to be greater than N, also we want B & N =0. Hence using these two facts B will be always (1<< (i+1)).
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; #define MAX 32 // Function to find A and B void findAB(int N) { bitset<MAX> arr(N), brr(N); // To store the leftmost set bit in N int leftsetN = -1; for (int i = MAX - 1; i >= 0; --i) { if (arr[i] == 1) { leftsetN = i; break; } } // To store the value of A int A = 0; for (int i = leftsetN; i >= 0; --i) { // If the bit is unset in N // then we will set it in A if (arr[i] == 0) { A |= (1 << i); } } // To store the value of B int B = 0; // B will be (1 << (leftsetN + 1)) B = 1 << (leftsetN + 1); // Print the values of A and B cout << "A = " << A << " and B = " << B; } // Driver code int main() { int N = 5; findAB(N); return 0; } |
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Output:
A = 2 and B = 8
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