HDOJ 1536 S-Nim

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3514    Accepted Submission(s): 1544

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

 

Source

Norgesmesterskapet 2004

 

Recommend

LL

 

 
人生中第一道SG函数题。。。。。

SG_DFS：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int k,s[111],t,m,a;
int sg[11000];

int SG_dfs(int x)
{
    if(sg[x]!=-1)
    {
        return sg[x];
    }
    int i;bool vis[111];
    memset(vis,false,sizeof(vis));
    for(i=0;s<=x&&i<k;i++)
    {
        SG_dfs(x-s);
        vis[sg[x-s]]=true;
    }
    for(i=0;i<=10000;i++)
    {
        if(!vis) break;
    }
    return sg[x]=i;
}

int main()
{
    while(scanf("%d",&k)!=EOF&&k)
    {
        for(int i=0;i<k;i++)
            scanf("%d",s+i);
        sort(s,s+k);
        memset(sg,-1,sizeof(sg));
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d",&m);
            int XOR=0;
            while(m--)
            {
                scanf("%d",&a);
                XOR^=SG_dfs(a);
            }
            printf("%c",XOR?'W':'L');
        }
        putchar(10);
    }
    return 0;
}

* This source code was highlighted by YcdoiT. ( style: Codeblocks )


SG打表：

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int k,s[111],t,m,a;
int sg[11000];

void getSG()
{
    memset(sg,0,sizeof(sg));
    bool flag[11000];
    for(int i=1;i<=10000;i++)
    {
        memset(flag,false,sizeof(flag));
        for(int j=0;j<k;j++)
        {
            if(s[j]>i) continue;
            flag[sg[i-s[j]]]=true;
        }
        for(int j=0;j<=10000;j++)
        {
            if(!flag[j])
            {
                sg=j;break;
            }
        }
    }
}

int main()
{
    while(scanf("%d",&k)!=EOF&&k)
    {
        for(int i=0;i<k;i++)
            scanf("%d",s+i);
        getSG();
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d",&m);
            int XOR=0;
            while(m--)
            {
                scanf("%d",&a);
                XOR^=sg[a];
            }
            printf("%c",XOR?'W':'L');
        }
        putchar(10);
    }
    return 0;
}

* This source code was highlighted by YcdoiT. ( style: Codeblocks )