S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1782 Accepted Submission(s): 802
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
//很郁闷，因为题目看不懂，只能看别人的代码。才知道意思……
//题意：有一个集合有S个数。玩N次取石子游戏，每次只能取S集合里的数个石子。求判先手胜负。
1 //递归写的
2 #include<iostream>
3 using namespace std;
4
5 int t;
6 int num[101];
7 int sg[10001];
8
9 int Get_sg(int n)
10 {
11 if(n==0) return 0;
12 int i;
13 bool mex[101]={0};
14 for(i=0;i<t;i++)
15 {
16 if(n>=num[i])
17 {
18 int temp=n-num[i];
19 if(sg[temp]==-1)
20 {
21 sg[temp]=Get_sg(temp);
22 }
23 mex[sg[temp]]=true;
24 }
25 }
26 for(i=0;;i++)
27 {
28 if(!mex[i])
29 {
30 return i;
31 }
32 }
33 }
34
35 int main()
36 {
37 int i,l,m,k;
38 while(scanf("%d",&t)!=EOF,t)
39 {
40 for(i=0;i<t;i++)
41 {
42 scanf("%d",&num[i]);
43 }
44 scanf("%d",&l);
45 memset(sg,-1,sizeof(sg));
46 while(l--)
47 {
48 int temp=0;
49 scanf("%d",&k);
50 while(k--)
51 {
52 scanf("%d",&m);
53 if(sg[m]==-1)
54 {
55 sg[m]=Get_sg(m);
56 }
57 temp^=sg[m];
58 }
59 if(temp)
60 {
61 printf("W");
62 }
63 else
64 {
65 printf("L");
66 }
67 }
68 putchar('\n');
69 }
70 return 0;
71 }
72