POJ S-Nim

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.

题目大意

 有l堆石子，游戏规则基本和nim游戏一样，但每次取得石子数只能是给定的数集S中的一个数。

输入有多组数据：

第一行第一个数k(0 < k ≤ 100)表示集合S有k个数，接下来k个数si(0 < si ≤ 10000) 。

第二行有一个数m(0 < m ≤ 100)，表示有m次游戏。

接下来m行，每行第一个数字l(0 < l ≤ 100)表示有l堆石子，接下来l个数字表示没每堆石子的个数。

输入以k=0结束

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

题解

SG函数的基础题，看懂题意即可。

#include<cstdio>#include<iostream>#include<cstring>#include<cstdlib>#include<cmath>using namespace std;int n,s[102],m;int sg[10002],ans;void dfs(int x){if(sg[x]!=-1) return;bool pd[10002];int y;memset(pd,false,sizeof(pd));for(int i=1;i<=n;i++)   {y=x-s[i];if (y>=0)   {if(sg[y]==-1) dfs(y);pd[sg[y]]=true;   }   }for(int i=0;i<=10000;i++){if(!pd[i])    {sg[x]=i;break;}}}void doit(int cs){int d,x;while(cs--)   {scanf("%d",&d);    ans=0;    for(int i=1;i<=d;i++)       {scanf("%d",&x);    if(sg[x]==-1) dfs(x);    ans=ans^sg[x];   }if(ans) printf("W");else printf("L");   }}int main(){while(scanf("%d",&n)&&n)   {for(int i=1;i<=n;i++) scanf("%d",&s[i]);    memset(sg,-1,sizeof(sg)); sg[0]=0;    scanf("%d",&m);    doit(m);    printf("\n");   }return 0;}