Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1232 Accepted Submission(s): 562
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 分析：一道非常不错的博弈题目，即考察了尼姆博弈又涉及了求sg值的问题。 可以将本题的代码作为求sg值的模版！具体代码如下： #include<cstdio> #include<algorithm> using namespace std; #define N 100+10 int knum,mnum,lnum; int ans[N],si[N],hi[N],sg[10010]; int mex(int x)//求x的sg值（可作为模版应用） { if(sg[x]!=-1) return sg[x]; bool vis[N];//注意要讲vis定义到函数里面！因为每次都要改变其值 memset(vis,false,sizeof(vis)); for(int i=0;i<knum;i++) { int temp=x-si[i]; if(temp<0) break; sg[temp]=mex(temp); vis[sg[temp]]=true; } for(int i=0;;i++) { if(!vis[i]) { sg[x]=i; break; } } return sg[x]; } int main() { freopen("game.in","r",stdin); freopen("game.out","w",stdout); while(scanf("%d",&knum) && knum) { for(int i=0;i<knum;i++) scanf("%d",&si[i]); sort(si,si+knum); memset(sg,-1,sizeof(sg)); sg[0]=0; memset(ans,0,sizeof(ans)); scanf("%d",&mnum); for(int i=0;i<mnum;i++) { scanf("%d",&lnum); for(int j=0;j<lnum;j++) { scanf("%d",&hi[i]); ans[i]^=mex(hi[i]);//尼姆博弈 } } for(int i=0;i<mnum;i++) { if(ans[i]==0) printf("L"); else printf("W"); } printf("/n"); } return 0; }
