S-Nim
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 rst 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
用dfs搜索sg值
1 #include <iostream> 2 #include <string.h> 3 #include <algorithm> 4 using namespace std; 5 int a[101]; 6 int t; 7 int b[10001]; 8 int dfs(int x) 9 { 10 if(b[x]!=-1)return b[x]; 11 int i; 12 if(x-a[0]<0)return 0; 13 int c[101]; 14 memset(c,0,sizeof(c)); 15 for(i=0;i<t;i++) 16 { 17 if(x-a[i]<0) 18 { 19 break; 20 } 21 c[dfs(x-a[i])]=1; 22 } 23 for(i=0;i<101;i++) 24 if(c[i]==0) 25 { 26 b[x]=i; 27 break; 28 } 29 return b[x]; 30 } 31 int main() 32 { 33 while(cin>>t&&t){ 34 memset(a,0,sizeof(a)); 35 memset(b,-1,sizeof(b)); 36 int j; 37 for(j=0;j<t;j++) 38 cin>>a[j]; 39 sort(a,a+t); 40 int i; 41 b[0]=0; 42 int n; 43 cin>>n; 44 for(i=0;i<n;i++) 45 { 46 int m; 47 int sum=0,x; 48 cin>>m; 49 for(j=0;j<m;j++) 50 { 51 cin>>x; 52 sum^=dfs(x); 53 } 54 if(sum) 55 cout<<"W"; 56 else cout<<"L"; 57 } 58 cout<<endl; 59 } 60 }