「POJ2960」S – Nim
Description
- 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
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
Print a newline after each test case.
Sample Input
1 2 3 4 5 6 7 8 9 10 11 |
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
1 2 |
LWW WWL |
Source
大意:有n堆石子,每堆石子个数已知,两人轮流从中取石子,
每次可取的石子数x满足x属于集合S(k) = {s1,s2,s3…sk},问先拿者是否有必胜策略?
分析:
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#include<iostream> #include<cstdio> #include<cstring> using namespace std; int k,s[101],sg[10001]; void get(int x) { if(sg[x]!=-1)return; bool mark[10001]; memset(mark,0,sizeof(mark)); int temp; for(int i=1;i<=k;i++) { temp=x-s[i]; if(temp>=0) { if(sg[temp]==-1) get(temp); mark[sg[temp]]=1; } } for(int i=0;;i++) if(!mark[i]) {sg[x]=i;return;} } int main() { int n,x,l; while(scanf("%d",&k)) { if(k==0)break; for(int i=1;i<=k;i++) scanf("%d",&s[i]); sg[0]=0; for(int i=1;i<=10000;i++)sg[i]=-1; scanf("%d",&n); while(n--) { scanf("%d",&l); int ans=0; while(l--) { scanf("%d",&x); if(sg[x]==-1)get(x); ans^=sg[x]; } if(ans)printf("W"); else printf("L"); } printf("\n"); } return 0; } |