제출: 11 통과: 3

[제출] [채점기록] [묻고답하기]

An N-omino is a two-dimensional shape formed by joining N unit cells fully along their edges in some way. More formally, a 1-omino is a 1x1 unit square, and an $N$-omino is an ($N-1$)omino with one or more of its edges joined to an adjacent 1x1 unit square. For the purpose of this problem, we consider two N-ominoes to be the same if one can be transformed into the other via reflection and/or rotation. For example, these are the five possible 4-ominoes:

And here are some of the 108 possible 7-ominoes:

Richard and Gabriel are going to play a game with the following rules, for some predetermined values of $X$, $R$, and $C$: 1. Richard will choose any one of the possible $X$-ominoes. 2. Gabriel must use at least one copy of that $X$-omino, along with arbitrarily many copies of any $X$-ominoes (which can include the one Richard chose), to completely fill in an $R$-by-$C$ grid, with no overlaps and no spillover. That is, every cell must be covered by exactly one of the $X$ cells making up an $X$-omino, and no $X$-omino can extend outside the grid. Gabriel is allowed to rotate or reflect as many of the $X$-ominoes as he wants, including the one Richard chose. If Gabriel can completely fill in the grid, he wins; otherwise, Richard wins. Given particular values $X$, $R$, and $C$, can Richard choose an $X$-omino that will ensure that he wins, or is Gabriel guaranteed to win no matter what Richard chooses?

The first line of the input gives the number of test cases, **T**.

**T** lines follow. Each contains three space-separated integers: **X**, **R**, and **C**.

**T** = 64.

1 ≤ **X, R, C** ≤ 4.

For each test case, output one line containing "Case #x: y", where $x$ is the test case number (starting from 1) and $y$ is either RICHARD (if there is at least one choice that ensures victory for Richard) or GABRIEL (if Gabriel will win no matter what Richard chooses).

```
4
2 2 2
2 1 3
4 4 1
3 2 3
```

```
Case #1: GABRIEL
Case #2: RICHARD
Case #3: RICHARD
Case #4: GABRIEL
```

In case #$1$, Richard only has one 2-omino available to choose -- the 1x2 block formed by joining two unit cells together. No matter how Gabriel places this block in the 2x2 grid, he will leave a hole that can be exactly filled with another 1x2 block. So Gabriel wins.

In case #$2$, Richard has to choose the 1x2 block, but no matter where Gabriel puts it, he will be left with a single 1x1 hole that he cannot fill using only 2-ominoes. So Richard wins.

In case #$3$, one winning strategy for Richard is to choose the 2x2 square 4-omino. There is no way for Gabriel to fit that square into the 4x1 grid such that it is completely contained within the grid, so Richard wins.

In case #$4$, Richard can either pick the straight 3-omino or the L-shaped 3-omino. In either case, Gabriel can fit it into the grid and then use another copy of the same 3-omino to fill in the remaining hole.