# SIMPLE_XOR - Editorial

Setter: Devendra Singh
Tester: Harris Leung
Editorialist: Trung Dang

1362

None

# PROBLEM:

You are given two integers L and R (L+3 \le R). Output any four distinct integers between L and R (inclusive) such that their bitwise XOR is zero.

If more than one such quadruple exists, you may output any of them. If no such quadruple exists, print β1 instead.

# EXPLANATION:

We have the following observation:

• If A is even, then A \oplus (A + 1) = 1.
• Therefore, if A even, then A \oplus (A + 1) \oplus (A + 2) \oplus (A + 3) = 1 \oplus 1 = 0.

With this observation, we have the following greedy algorithm:

• Try whether L \oplus (L + 1) \oplus (L + 2) \oplus (L + 3) = 0 or not. If yes, output these four numbers.
• If not, there are two cases:
• If R = L + 3, then there canβt be any other four distinct numbers between L and R. We output -1.
• Else, we know that L must be odd. Then, we know that (L + 1) \oplus (L + 2) \oplus (L + 3) \oplus (L + 4) = 0, so we output these four numbers.

# TIME COMPLEXITY:

Time complexity is O(1) per test case.

# SOLUTION:

Setter's Solution
``````#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define pb push_back
#define all(_obj) _obj.begin(), _obj.end()
#define F first
#define S second
#define pll pair<ll, ll>
#define vll vector<ll>
ll INF = 1e18;
const int N = 1e5 + 11, mod = 1e9 + 7;
ll max(ll a, ll b) { return ((a > b) ? a : b); }
ll min(ll a, ll b) { return ((a > b) ? b : a); }
void sol(void)
{
int l, r;
cin >> l >> r;
if ((r - l) > 3 || l % 2 == 0)
{
if (l % 2 == 0)
cout << l << ' ' << l + 1 << ' ' << l + 2 << ' ' << l + 3 << '\n';
else
cout << l + 1 << ' ' << l + 2 << ' ' << l + 3 << ' ' << l + 4 << '\n';
}
else
cout << -1 << '\n';
return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
int test = 1;
cin >> test;
while (test--)
sol();
}
``````
Tester's Solution
``````#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define fi first
#define se second
const ll mod=998244353;
const int N=2e6+1;
int main(){
ios::sync_with_stdio(false);
int t;cin >> t;
while(t--){
ll l,r;
cin >> l >> r;
if(l%2==0) cout << l << ' ' << l+1 << ' ' << l+2 << ' ' << l+3 << '\n';
else if(r==l+3) cout << "-1\n";
else{
l++;
cout << l << ' ' << l+1 << ' ' << l+2 << ' ' << l+3 << '\n';
}
}
}
``````
Editorialist's Solution
``````#include <bits/stdc++.h>
using namespace std;

int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int t; cin >> t;
while (t--) {
int l, r; cin >> l >> r;
if ((l ^ (l + 1) ^ (l + 2) ^ (l + 3)) == 0) {
cout << l << " " << l + 1 << " " << l + 2 << " " << l + 3 << '\n';
} else if (l + 4 <= r) {
cout << l + 1 << " " << l + 2 << " " << l + 3 << " " << l + 4 << '\n';
} else {
cout << "-1\n";
}
}
}
``````
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