hdu 1536&&1944 S-Nim sg函数 难度:0

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4770    Accepted Submission(s): 2058

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

 

感想:这两道真是同一题

思路:sg硬打表

 

#include <cstdio>

#include <cstring>

#include <algorithm>

using namespace std;

int s[101],k;

int sg[10001];

int n;

int seg(int num){

    if(sg[num]!=-1)return sg[num];

    bool used[101];

    memset(used,0,sizeof(used));

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

        if(num<s[i])break;

        used[seg(num-s[i])]=true;

    }

    for(int i=0;i<101;i++)if(!used[i])return sg[num]=i;

}

int main(){

    freopen("C:\\Users\\Administrator\\Desktop\\input.txt","w",stdout);

    while(scanf("%d",&k)==1&&k){

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

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

        sort(s,s+k);

        int T;

        scanf("%d",&T);

        while(T--){

            scanf("%d",&n);

            int sum=0,tmp;

            for(int i=0;i<n;i++){scanf("%d",&tmp);sum^=seg(tmp);}

            if(sum==0)putchar('L');

            else putchar('W');

        }

        putchar('\n');

    }

    return 0;

}