Competitive Programming Solutions

1472B - Fair Division

04 Jan 2021 — Tags: math — URL

First, if the total sum is not even, then it’s impossible. Assume otherwise. Second, if we have an even amonut of $2$’s (we’ll necesarily have an even number of $1$s) we can split it into two equal groups with the same sum. Otherwise, we need at least a pair of $1$’s to fill the missing $2$.

Time complexity: $O(n)$

Memory complexity: $O(n)$

Click to show code.

 

using namespace std;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
bool solve(vi a)
{
    int sum = accumulate(begin(a), end(a), 0), n = (int)(a).size();
    int ones = count_if(begin(a), end(a), [](int x) { return x == 1; });
    int twos = n - ones;
    return sum % 2 == 0 and ((twos % 2 == 0) or (ones > 0));
}
int main(void)
{
    ios::sync_with_stdio(false), cin.tie(NULL);
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        vi a(n);
        for (auto &ai : a)
            cin >> ai;
        cout << (solve(a) ? "YES" : "NO") << endl;
    }
    return 0;
}

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