hdu 1536 S-Nim

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S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8570    Accepted Submission(s): 3595


Problem Description

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 5
3
2 5 12
3 2 4 7
4 2 3 7 12
5 1 2 3 4 5
3
2 5 12
3 2 4 7
4 2 3 7 12
0

 

Sample Output

LWW
WWL

思路：n堆石子，分成n个游戏，结果是所有游戏的结合。

需要用到SG函数，百度百科的解释

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
int s[110];
int sg[10010];
int k;
int SG_dfs(int x)
{
int i,e;
if(sg[x]!=-1)
return sg[x];
bool vis[110];
memset(vis,0,sizeof(vis));
for(i=0;i<k;i++)
{
if(x>=s[i])
{
SG_dfs(x-s[i]);
vis[sg[x-s[i]]]=1;
}
}
for(i=0;;i++)
{
if(!vis[i])
{
e=i;
break;
}
}
return sg[x]=e;
}
int main()
{
int m,num,t;
while(cin>>k,k)
{
for(int i=0;i<k;i++)
cin>>s[i];
memset(sg,-1,sizeof(sg));
sort(s,s+k);
cin>>m;
while(m--)
{
cin>>t;
int ans=0;
while(t--)
{
cin>>num;
ans^=SG_dfs(num);
}
if(ans==0)
printf("L");
else
printf("W");
}
printf("\n");
}
return 0;
}