1455C - Ping Pong

Keep in mind that the first priority of each player is to maximize their wins.

Let’s take the perspective of player $2$, Bob. Bob can win at most $y$, times, by doing nothing and letting player $1$ run out of stamina. Can Bob minimize Alice’s wins without affecting his score?

One way is to only respond on Alice’s last serve (since she wouldn’t be able to respond anyway),. But we soon realize that’s the only option.

If Bob responds to any of Alice’s $i \in {1, 2, …, x - 1}$ serve, then Alice’s best chance is to not respond until Bob’s last serve, $y$ (same strategy). Alice would then win $(i - 1) + (x - i) = x - 1$ times and Bob $y-1$ times. This is suboptimal for Bob, since we know he can win $y$ times by our optimal strategy.

Since Alice is on Bob’s mercy (she can’t stop serving until she runs out of stamina or Bob wins), Bob will choose the optimal strategy and win $y$ times while Alice $x - 1$ times.

Time complexity: $O(1)$

Memory complexity: $O(1)$

using namespace std;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
int main(void)
{
    ios::sync_with_stdio(false), cin.tie(NULL);
    int t;
    cin >> t;
    while (t--)
    {
        int x, y;
        cin >> x >> y;
        cout << x - 1 << " " << y << endl;
    }
    return 0;
}