Game

Research check: the canonical statement is the Codeforces problem page for 2199A - Game, and the official Kotlin Heroes 14 tutorial page lists the reference solution condition as a1 - b1 + a2 - b2 >= k ? NO : YES on Codeforces. The tutorial text itself was basically empty, so the real work is proving why that tiny condition is not a trap.

Let

$D=(a_1-b_1)+(a_2-b_2).$

This is Alice's total score minus Bob's total score after the first two rounds. If $D$ is large, Alice is ahead; if $D$ is negative, Bob is already ahead.

Bob wants to know whether there exists some valid third round that makes him win the game. Since third-round scores must be between $0$ and $k$, Bob's best possible score swing is achieved by

$a_3=0,\qquad b_3=k.$

This is always legal because $k\ge 1$, so $a_3\ne b_3$. It decreases Alice's lead by exactly $k$. After this best-case round for Bob, the final score difference becomes

$D-k.$

If $D<k$, then $D-k<0$, so Bob's total score is strictly larger. That immediately means Bob wins. No tiebreaker needed.

Now suppose $D>k$. Even Bob's maximum possible swing leaves

$D-k>0,$

so Alice still has the larger total. Bob is cooked.

The only slightly annoying case is $D=k$. Then Bob can tie the total score by playing $(a_3,b_3)=(0,k)$. Could he win by having more round wins? No.

Why? Each of the first two round differences is at most $k$:

$a_i-b_i\le k.$

Also, every round has unequal scores, so a round Bob wins contributes a negative value to $D$. If Bob had won either of the first two rounds, one term would be negative, and the other term could be at most $k$, making the sum strictly less than $k$. Therefore $D=k$ implies Alice won both first two rounds. In the third round Bob can win one round, but the round count becomes Alice $2$, Bob $1$, so Alice still wins the tiebreaker.

So Alice definitely wins exactly when $D\ge k$. Bob has a chance exactly when

$D<k.$

Implementation is just compute $D$ and print YES if $D<k$, otherwise NO.

Complexity per test case is $O(1)$ time and $O(1)$ memory.