# EQUALDIST-Editorial

Setter: Lavish Gupta
Tester: Nishank Suresh, Abhinav sharma
Editorialist: Devendra Singh

330

None

# PROBLEM:

Alice and Bob are very good friends, and they always distribute all the eatables equally among themselves.

Alice has A chocolates, and Bob has B chocolates. Determine whether Alice and Bob can distribute all the chocolates equally among themselves.

Note that it is not allowed to break a chocolate into more than one piece.

# EXPLANATION:

If A+B is even, both of them can have \frac{A+B}{2} chocolates otherwise both of them cannot have equal number of chocolates as the total number of chocolates is odd.

# TIME COMPLEXITY:

O(1) for each test case.

# SOLUTION:

Editorialist'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 a, b;
cin >> a >> b;
cout << (((a + b) % 2) ? "NO\n" : "YES\n");
return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
int test = 1;
cin >> test;
while (test--)
sol();
}


1 Like

i did like if(difference(a,b) % 2 == 0) then return YES, or else return NO,
And it worked, i really dont know how XD