POJ-2960-S-Nim（SG函数）

题目链接：http://poj.org/problem?id=2960

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S-Nim

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 3880 Accepted: 2034

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 2 532 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

Source

Svenskt Mästerskap i Programmering/Norgesmesterskapet 2004

SG函数详解：http://www.cnitblog.com/weiweibbs/articles/42735.html

//#include <bits/stdc++.h>#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int Scan(){    int res=0,ch,flag=0;    if((ch=getchar())=='-')flag=1;    else if(ch>='0'&&ch<='9')res=ch-'0';    while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-'0';    return flag?-res:res;}void Out(int a){    if(a>9)Out(a/10);    putchar(a%10+'0');}#define INF 0x3f3f3f3f#define bug cout<<"bug"<<endlconst int MAXN = 1e4+7;int a[MAXN],poi[MAXN];int sg[MAXN];int n,m,k,ans,p;bool vis[MAXN];void get_sg(){    for(int i=0; i<MAXN; ++i)    {        memset(vis,0,sizeof(vis));        for(int j=0; j<n && poi[j]<=i; ++j)            vis[ sg[i-poi[j]] ]=1;        for(int j=0; ;++j)if(!vis[j])        {sg[i]=j;break;}    }}int main(){    while(scanf("%d",&n)!=EOF)    {        if(!n)break;        for(int i=0; i<n; ++i)            poi[i]=Scan();        sort(poi,poi+n);        get_sg();        m=Scan();        for(int i=0; i<m; ++i)        {            k=Scan(),ans=0;            for(int j=0; j<k; ++j)            {                p=Scan();                ans^=sg[p];            }            if(ans)putchar('W');            else putchar('L');        }        putchar('\n');    }    return 0;}