- 描述
- Arthur and his sister Caroll have been playing a game called Nim for some time now. Nim is played as follows:
- The starting position has a number of heaps, all containing some, not necessarily equal, number of beads.
- The players take turns chosing a heap and removing a positive number of beads from it.
- The first player not able to make a move, loses.
Arthur and Caroll really enjoyed playing this simple game until they
recently learned an easy way to always be able to find the best move:- Xor the number of beads in the heaps in the current position (i.e. if we have 2, 4 and 7 the xor-sum will be 1 as 2 xor 4 xor 7 = 1).
- If the xor-sum is 0, too bad, you will lose.
- Otherwise, move such that the xor-sum becomes 0. This is always possible.
It is quite easy to convince oneself that this works. Consider these facts:- The player that takes the last bead wins.
- After the winning player's last move the xor-sum will be 0.
- The xor-sum will change after every move.
Which means that if you make sure that the xor-sum always is 0 when you have made your move, your opponent will never be able to win, and, thus, you will win.
Understandibly it is no fun to play a game when both players know how to play perfectly (ignorance is bliss). Fourtunately, Arthur and Caroll soon came up with a similar game, S-Nim, that seemed to solve this problem. Each player is now only allowed to remove a number of beads in some predefined set S, e.g. if we have S = f2, 5g each player is only allowed to remove 2 or 5 beads. Now it is not always possible to make the xor-sum 0 and, thus, the strategy above is useless. Or is it?
your job is to write a program that determines if a position of S-Nim is a losing or a winning position. A position is a winning position if there is at least one move to a losing position. A position is a losing position if there are no moves to a losing position. This means, as expected, that a position with no legal moves is a losing position.c
- 输入
- Arthur and his sister Caroll have been playing a game called Nim for some time now. Nim is played as follows:
- The starting position has a number of heaps, all containing some, not necessarily equal, number of beads.
- The players take turns chosing a heap and removing a positive number of beads from it.
- The first player not able to make a move, loses.
Arthur and Caroll really enjoyed playing this simple game until they
recently learned an easy way to always be able to find the best move:- Xor the number of beads in the heaps in the current position (i.e. if we have 2, 4 and 7 the xor-sum will be 1 as 2 xor 4 xor 7 = 1).
- If the xor-sum is 0, too bad, you will lose.
- Otherwise, move such that the xor-sum becomes 0. This is always possible.
It is quite easy to convince oneself that this works. Consider these facts:- The player that takes the last bead wins.
- After the winning player's last move the xor-sum will be 0.
- The xor-sum will change after every move.
Which means that if you make sure that the xor-sum always is 0 when you have made your move, your opponent will never be able to win, and, thus, you will win.
Understandibly it is no fun to play a game when both players know how to play perfectly (ignorance is bliss). Fourtunately, Arthur and Caroll soon came up with a similar game, S-Nim, that seemed to solve this problem. Each player is now only allowed to remove a number of beads in some predefined set S, e.g. if we have S = f2, 5g each player is only allowed to remove 2 or 5 beads. Now it is not always possible to make the xor-sum 0 and, thus, the strategy above is useless. Or is it?
your job is to write a program that determines if a position of S-Nim is a losing or a winning position. A position is a winning position if there is at least one move to a losing position. A position is a losing position if there are no moves to a losing position. This means, as expected, that a position with no legal moves is a losing position.
- 输出
- For each position: If the described position is a winning position print a 'W'.If the described position is a losing position print an 'L'.
Print a newline after each test case.c
- 样例输入
For each position: If the described position is a winning position print a 'W'.If the described position is a losing position print an 'L'.
Print a newline after each test case.
- 样例输出
LWW
WWL
