S-Nim
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 3356 | Accepted: 1769 |
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
Source
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<cstdlib> #include<string> #include<algorithm> using namespace std; int num[105],k,m,n,res; int sg[10010]; int SG(int x) { int i; if(sg[x]+1) return sg[x]; bool vis[10010]={0}; for(i=1;i<=k;i++) { if(x>=num[i]) vis[SG(x-num[i])]=1; } for(i=0;vis[i];i++); return sg[x]=i; } int main() { while(scanf("%d",&k)!=EOF) { if(k==0) break; for(int i=1;i<=k;i++) scanf("%d",&num[i]); memset(sg,-1,sizeof(sg)); sg[0]=0; scanf("%d",&m); while(m--) { int nim=0; scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d",&res); nim=nim^SG(res); } if(nim) printf("W"); else printf("L"); } printf("\n"); } return 0; }