ACM-SG函數之S-Nim——hdu1536 hdu1944 poj2960

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4091    Accepted Submission(s): 1760

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   題目：http://acm.hdu.edu.cn/showproblem.php?pid=1536

這也是一道經典SG函數的題目。 有關于SG函數的解，可以戳這個，很詳細→http://blog.csdn.net/lttree/article/details/24886205 這道題題意： 我就按着樣例格式來說吧： 先輸入一個K，表示取數集合的個數。（K為0，則結束） 後面跟k個數,表示取數集合的數（就是每次隻能取這幾個數量的物品） 然後會跟一個M,表示有M次詢問。 然後接下來M行，每行先有一個N，表示有多少堆物品。 N後跟着N個數，表示每堆物品數量。

因為，OJ背景的操作，輸入和輸出是分開的（其實就是将你的程式的答案存成一個TXT檔案，然後和 标準答案TXT檔案進行二進制的比較） 是以，我每個N都直接輸出'L'或者'W‘， 在M行結束時，換行，沒有用數組來存答案。 PS：用scanf比cin快80MS

/************************************************
*************************************************
*        Author:Tree                            *
*From :http://blog.csdn.net/lttree              *
* Title : S-Nim                                 *
*Source: hdu 1536                               *
* Hint  : SG                                    *
*************************************************
*************************************************/
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define N 10001
int f[N],sg[N];
bool mex[N];
void get_sg(int t,int n)
{
    int i,j;
    memset(sg,0,sizeof(sg));
    for(i=1;i<=n;i++)
    {
        memset(mex,0,sizeof(mex));
        // 對于屬于g(x)後繼的數置1
        for( j=1 ;j<=t && f[j]<=i ;j++ )
            mex[sg[i-f[j]]]=1;
        // 找到最小不屬于該集合的數
        for( j=0 ; j<=n ; j++ )
            if(!mex[j])
                break;
        sg[i] = j;
    }
}
int main()
{
    int k,m,n,i,t,temp;
    while( scanf("%d",&k) && k )
    {
        for(i=1;i<=k;++i)
            scanf("%d",&f[i]);
        sort(f+1,f+k+1);
        get_sg(k,N);
        scanf("%d",&m);
        while(m--)
        {
            temp=0;
            scanf("%d",&n);
            for(i=0;i<n;++i)
            {
                scanf("%d",&t);
                temp^=sg[t];
            }
            if( !temp )  printf("L");
            else    printf("W");
        }
        printf("\n");
    }
    return 0;
}