HDOJ 题目S-Nim（sg-博弈，模板）

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4522    Accepted Submission(s): 1959

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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 ac代码

#include<stdio.h>#include<stdlib.h>#include<string.h>int n,s[10100],sg[11000];//int v[100000];int cmp(const void *a,const void *b){    return *(int *)a-*(int *)b;}int dfs(int x){    int i,e;    int v[110];//应该定义到里边    if(sg[x]!=-1) return sg[x];    memset(v,0,sizeof(v));    for(i=0;i<n;i++)    {        if(x>=s[i])        {            dfs(x-s[i]);            v[sg[x-s[i]]]=1;        }    }    for(i=0;;i++)    {        if(!v[i])        {            e=i;            break;        }    }    return sg[x]=e;}int main(){    int i,t;    while(scanf("%d",&n)!=EOF,n)    {        memset(sg,-1,sizeof(sg));        for(i=0;i<n;i++)            scanf("%d",&s[i]);        qsort(s,n,sizeof(s[0]),cmp);        /*for(i=0;i<n;i++)            printf("%d ",s[i]);*/    //    printf("\n");        scanf("%d",&t);        while(t--)        {            int c,sum=0;                        scanf("%d",&c);            while(c--)            {                int num;                scanf("%d",&num);                sum^=dfs(num);            //    printf("sg[%d]=%d\n",num,sg[num]);            }            if(sum==0)                printf("L");            else                printf("W");        }        printf("\n");    }}