hdu 1536 SG函数模板题

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3077    Accepted Submission(s): 1361

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

 

题意：

首先输入K 表示一个集合的大小  之后输入集合 表示对于这对石子只能去这个集合中的元素的个数

之后输入 一个m 表示接下来对于这个集合要进行m次询问 

之后m行 每行输入一个n 表示有n个堆  每堆有n1个石子  问这一行所表示的状态是赢还是输 如果赢输入W否则L

 

思路：

对于n堆石子 可以分成n个游戏 之后把n个游戏合起来就好了

 

#include<stdio.h>

#include<string.h>

#include<algorithm>

using namespace std;

//注意 S数组要按从小到大排序 SG函数要初始化为-1 对于每个集合只需初始化1边

//不需要每求一个数的SG就初始化一边

int SG[10100],n,m,s[102],k;//k是集合s的大小 S[i]是定义的特殊取法规则的数组

int dfs(int x)//求SG[x]模板

{

    if(SG[x]!=-1) return SG[x];

    bool vis[110];

    memset(vis,0,sizeof(vis));



    for(int i=0;i<k;i++)

    {

        if(x>=s[i])

        {

           dfs(x-s[i]);

           vis[SG[x-s[i]]]=1;

         }

    }

    int e;

    for(int i=0;;i++)

      if(!vis[i])

      {

        e=i;

        break;

      }

    return SG[x]=e;

}

int main()

{

    int cas,i;

    while(scanf("%d",&k)!=EOF)

    {

        if(!k) break;

        memset(SG,-1,sizeof(SG));

        for(i=0;i<k;i++) scanf("%d",&s[i]);

        sort(s,s+k);

        scanf("%d",&cas);

        while(cas--)

        {

            int t,sum=0;

            scanf("%d",&t);

            while(t--)

            {

                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");

    }

    return 0;

}

 

 下面是对SG打表的做法

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

const int K=101;

const int H=10001;//H是我们要打表打到的最大值

int k,m,l,h,s[K],sg[H],mex[K];///k是集合元素的个数 s[]是集合  mex大小大约和集合大小差不多

///注意s的排序

void sprague_grundy()

{

    int i,j;

    sg[0]=0;

    for (i=1;i<H;i++){

        memset(mex,0,sizeof(mex));

        j=1;

        while (j<=k && i>=s[j]){ 

            mex[sg[i-s[j]]]=1;

            j++;

        }

        j=0;

        while (mex[j]) j++;

        sg[i]=j;

    }

}



int main(){

    int tmp,i,j;



    scanf("%d",&k);

    while (k!=0){

        for (i=1;i<=k;i++)

            scanf("%d",&s[i]);

        sort(s+1,s+k+1);            //这个不能少

        sprague_grundy();

        scanf("%d",&m);

        for (i=0;i<m;i++){

            scanf("%d",&l);

            tmp=0;

            for (j=0;j<l;j++){

                scanf("%d",&h);

                tmp=tmp^sg[h];

            }

            if (tmp)

                putchar('W');

            else

                putchar('L');

        }

        putchar('\n');

        scanf("%d",&k);

    }

    return 0;}