S-Nim（hdu1536+SG函數）

S-Nim

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5317 Accepted Submission(s): 2288

 

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

 

 

題意：首先輸入K 表示一個集合的大小 之後輸入集合 表示對於這對石子只能去這個集合中的元素的個數

之後輸入 一個m 表示接下來對於這個集合要進行m次詢問

之後m行 每行輸入一個n 表示有n個堆 每堆有n1個石子 問這一行所表示的狀態是贏還是輸 如果贏輸入W否則L

思路：對於n堆石子 可以分成n個游戲 之後把n個游戲合起來就好了

 

轉載請注明出處：尋找&星空の孩子

 

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

 

#include
#include
#include
using namespace std;
//注意 S數組要按從小到大排序 SG函數要初始化為-1 對於每個集合只需初始化1遍
//n是集合s的大小 S[i]是定義的特殊取法規則的數組
int s[110],sg[10010],n;
int SG_dfs(int x)
{
    int i;
    if(sg[x]!=-1)
        return sg[x];
    bool vis[110];
    memset(vis,0,sizeof(vis));
    for(i=0;i=s[i])
        {
            SG_dfs(x-s[i]);
            vis[sg[x-s[i]]]=1;
        }
    }
    int e;
    for(i=0;;i++)
        if(!vis[i])
        {
            e=i;
            break;
        }
    return sg[x]=e;
}
int main()
{
    int i,m,t,num;
    while(scanf(%d,&n)&&n)
    {
        for(i=0;i