S-nim
| Time Limit: 2000MS |
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Memory Limit: 65536K |
| Total Submissions: 3356 |
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Accepted: 1769 |
Description
Arthur and his sister Caroll has been playing a game called Nim for some time now. Nim is played as follows:
- The starting position have 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 isn't able to make a move, loses.
Arthur and Caroll really enjoyed playing this simple game until they
Recently learned an easy-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 would be 1 as 2 XOR 4 XOR 7 = 1).
- If The xor-sum is 0, too bad, you'll lose.
- Otherwise, move such that the xor-sum becomes 0. This was always possible.
It's quite easy-to-convince oneself that's this works. Consider these facts:
- The player is takes the last bead wins.
- After the winning player's last move the xor-sum would be 0.
- The xor-sum would change after every move.
Which means that if you do sure that the xor-sum always was 0 when you had made your move, your opponent would never be a Ble to win, and, thus, you'll win.
Understandibly It is no 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, which seemed to solve this problem. Each player was now only allowed to remove a number of beads in some predefined set S, e.g. if we had s = {2, 5} each play Er is only allowed to remove 2 or 5 beads. Now it isn't 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 this determines if a position of S-nim is a losing or a winning position. A position is a winning position if there are at least one move to a losing position. A position is a losing position if there be no moves to a losing position. This means, as expected, which 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 was 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 a ' L '.
Print a newline after all test case.
Sample Input
2 2 532 5 123 2 4 74 2 3 7 125 1 2 3 4 532 5 123 2 4 74 2 3 7 120
Sample Output
Lwwwwl
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;}
(SG) POJ 2960
