poj 2960 S-Nim(SG函数)
2016-01-30 20:08
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S-Nim
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
Sample Output
Source
Svenskt Mästerskap i Programmering/Norgesmesterskapet 2004
【思路】
SG函数。
裸ti ,注意下sg和vis的大小就好了 :)
【代码】
Time Limit: 2000MS | Memory Limit: 65536K | |
Total Submissions: 3694 | Accepted: 1936 |
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
Svenskt Mästerskap i Programmering/Norgesmesterskapet 2004
【思路】
SG函数。
裸ti ,注意下sg和vis的大小就好了 :)
【代码】
#include<cstdio> #include<cstring> #define FOR(a,b,c) for(int a=(b);a<(c);a++) using namespace std; int n,m,a[101],sg[10001]; int dfs(int x) { if(sg[x]!=-1) return sg[x]; if(!x) return sg[x]=0; int vis[10001]; //size of [si] memset(vis,0,sizeof(vis)); FOR(i,0,n) if(x>=a[i]) vis[dfs(x-a[i])]=1; for(int i=0;;i++) if(!vis[i]) return sg[x]=i; } int main() { while(scanf("%d",&n)==1 && n) { FOR(i,0,n) scanf("%d",&a[i]); scanf("%d",&m); memset(sg,-1,sizeof(sg)); FOR(i,0,m) { int x,v,ans=0; scanf("%d",&x); FOR(j,0,x) scanf("%d",&v) , ans^=dfs(v); if(ans) printf("W"); else printf("L"); } putchar('\n'); } return 0; }
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