CHFNA - Editorial

PROBLEM LINK:

Practice
Contest

Author: Deepak Chaudhary
Tester: Vikas Yadav
Editorialist: Deepak Chaudhary

DIFFICULTY:

EASY

PREREQUISITES:

Bitwise-XOR

PROBLEM:

Find A for given two numbers X and Y such that (A⊕X)+(A⊕Y) will give minimum value.

QUICK EXPLANATION:

For minimum value (A⊕X)+(A⊕Y) can be written as (X ⊕ Y).

EXPLANATION:

Think about addition in base two. Say X = 10101 and Y = 1001. What your operation does is it modifies the bits in your numbers, so if the first bit in X is 1 and the first bit in Y is 1 (as is the case above) you can make both 0 by making that bit 1 in A. This is actually the only way you can decrease the resulting sum, so A = 1 is an answer above.

SOLUTIONS:

Setter's Solution
// Chef and Chefina

#include <bits/stdc++.h>
using namespace std;
#define int long long

void solve()
{
    int test_cases;
    cin >> test_cases;

    while(test_cases--)
    {
        int a, b;
        cin >> a >> b;
        cout << (a ^ b) << "\n";
    }
}

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    solve();

    return 0;
}

Tester's Solution
// Chef and Chefina

#include <stdio.h>

int main()
{
    int test_cases;
    scanf("%d", &test_cases);

    while(test_cases--)
    {
        int x, y;
        scanf("%d %d", &x, &y);
        printf("%ld\n", x ^ y);
    }

    return 0;
}