HDOJ S-Nim 1536&POJ S-Nim 2960【求SG函数+Nim游戏】

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5288    Accepted Submission(s): 2275

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 2 532 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

Norgesmesterskapet 2004

 

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主要是求SG函数 然后利用Nim游戏求解方法求解 由此种游戏称Nim和

称一个游戏G 是一组游戏G1 G2 。。。Gn的和 G的每一步可抽象成：先选一个子游戏Gi  然后在Gi 中走一步。

如果gi是Gi的SG函数，则G的SG函数 g=g1^g2^g3^...^gn  对于G任意状态a=(a1,a2,,...,an),g(a)=g(a1)^g(a2)^...^g(an)。

此题将每次可取的数看成一个Gi 然后求解gi。最后进行Nim游戏。

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int sg[10010];int s[110];int k,m,l;int mex(int x){    if(sg[x]!=-1)return sg[x];    bool vis[110]; //一定要将vis数组定义到这里    memset(vis,false,sizeof(vis));    for(int i=0;i<k;i++){        int temp=x-s[i];        if(temp<0)break;//此处因为s[i]排好序的，所以后面temp的值肯定小于0，所以直接跳出就可以。        sg[temp]=mex(temp);        vis[sg[temp]]=true;    }    for(int i=0;;i++){        if(!vis[i]){            sg[x]=i;            break;        }    }    return sg[x];}int main(){    while(scanf("%d",&k),k)    {        int res[110];        memset(res,0,sizeof(res));        memset(sg,-1,sizeof(sg));        sg[0]=0;        for(int i=0;i<k;i++)            scanf("%d",&s[i]);        sort(s,s+k);        scanf("%d",&m);        int a;        for(int i=0;i<m;i++){            scanf("%d",&l);            for(int j=0;j<l;j++){                scanf("%d",&a);                res[i]^=mex(a);            }        }        for(int i=0;i<m;i++){            if(res[i]) printf("W");            else printf("L");        }        printf("\n");    }    return 0;}