(简单) 博弈 HOJ 2128 S-Nim

S-Nim

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Source : Nordic Collegiate Programming Contest 2004Time limit : 3 secMemory limit : 32 M

Submitted : 80, Accepted : 50

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 = {2, 5} 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.

Input

Input consists of a number of test cases.

For each test case: The first line contains a number k (0 < k <= 100) describing the size of S, followed by k numbers si (0 < si <= 10000) describing S. The second line contains a number m (0 < m <= 100) describing the number of positions to evaluate. The next m lines each contain a number l (0 < l <= 100) describing the number of heaps and l numbers hi (0 <= hi <= 10000) describing the number of beads in the heaps.

The last test case is followed by a 0 on a line of its own.

Output

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.

Sample Input

2 2 532 5 123 2 4 74 2 3 7 125 1 2 3 4 532 5 123 2 4 74 2 3 7 120

Sample Output

LWWWWL

题意：有n堆石子，两个人轮流选一堆拿一定数量的石子，其中的数量是S集合中的任意一个数。问假设两个人都足够聪明，最终谁会获胜。

思路：这题就是Nim游戏的变形，我们用sg[x] 表示一堆石子中有x个石子的状态，sg[x]的运算方法，假设从x状态能到达yi 状态 ， 那么sg[x] = min ( 自然数 ， 且!=sg[yi] )  yi为任意一个x能到达的状态。

那么最后的答案就是每堆石子对应的sg值异或一下 ， 如果最后不为0，则先手必胜，否则必败。

代码：

#include<iostream>

#include<stdio.h>

#include<math.h>

#include<cstring>

#include<algorithm>

using namespace std;

#pragma comment (linker , "/STACK:10240000000000")

int S[110];

int sg[10010];

int k , V;

int dfs(int x)

{

if (sg[x]!=-1) return sg[x];

bool *vis = new bool [V+10];  // 不用动态分配内存会爆栈

for (int i = 0 ; i <= V ; ++i) vis[i] = false;

for (int i = 0 ; i < k && x-S[i] >= 0 ; ++i)

{

vis[dfs(x-S[i])] = true;

}

for (int i = 0 ; i <= V ; ++i) if (!vis[i])

{

delete [ ] vis;

return sg[x] = i;

}

}

int main()

{

while (scanf("%d",&k) , k)

{

for (int i = 0 ; i < k ; ++i) scanf("%d",S+i);

memset(sg,-1,sizeof(sg));

sg[0] = 0;

sort(S,S+k);

int T;

scanf("%d",&T);

while (T--)

{

int n;

scanf("%d",&n);

V = 0;

int ans = 0;

for (int i = 0 ; i < n ; ++i)

{

int tmp;

scanf("%d",&tmp);

if (tmp>V) V = tmp;

ans ^= dfs(tmp);

}

if (ans) printf("W");

else printf("L");

}

printf("\n");

}

}