S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4527 Accepted Submission(s): 1960
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
记忆化求sg，直接打表会tle
1 #include<iostream>
2 #include<string>
3 #include<cstdio>
4 #include<vector>
5 #include<queue>
6 #include<stack>
7 #include<set>
8 #include<algorithm>
9 #include<cstring>
10 #include<stdlib.h>
11 #include<math.h>
12 using namespace std;
13 #define ll __int64
14 int sg[11000],ss[110],k;
15 int getsg(int n){
16 int vit[110];
17 memset(vit,0,sizeof(vit));
18 for(int j=0;j<k&&ss[j]<=n;j++){
19 if(sg[n-ss[j]]==-1) sg[n-ss[j]]=getsg(n-ss[j]);
20 vit[sg[n-ss[j]]]=1;
21 }
22 for(int j=0;;j++)
23 if(!vit[j]) return j;
24 }
25 int main(){
26 while(scanf("%d",&k)&&k){
27 for(int i=0;i<k;i++) scanf("%d",&ss[i]);
28 sort(ss,ss+k);
29 int m;scanf("%d",&m);
30 memset(sg,-1,sizeof(sg));
31 sg[0]=0;
32 while(m--){
33 int ans=0,t;scanf("%d",&t);
34 while(t--){
35 int tt;scanf("%d",&tt);
36 if(sg[tt]==-1) sg[tt]=getsg(tt);
37 ans^=sg[tt];
38 }
39 if(ans==0) printf("L");
40 else printf("W");
41 }
42 printf("\n");
43 }
44 }