ACM-SG函数之S-Nim——hdu1536 hdu1944 poj2960
S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4091 Accepted Submission(s): 1760
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.
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
Norgesmesterskapet 2004
题目：http://acm.hdu.edu.cn/showproblem.php?pid=1536
这也是一道经典SG函数的题目。
有关于SG函数的解，可以戳这个，很详细→http://blog.csdn.net/lttree/article/details/24886205
这道题题意：
我就按着样例格式来说吧：
先输入一个K，表示取数集合的个数。（K为0，则结束）
后面跟k个数,表示取数集合的数（就是每次只能取这几个数量的物品）
然后会跟一个M,表示有M次询问。
然后接下来M行，每行先有一个N，表示有多少堆物品。
N后跟着N个数，表示每堆物品数量。
因为，OJ后台的操作，输入和输出是分开的（其实就是将你的程序的答案存成一个TXT文件，然后和
标准答案TXT文件进行二进制的比较）
所以，我每个N都直接输出'L'或者'W‘，
在M行结束时，换行，没有用数组来存答案。
PS：用scanf比cin快80MS
/************************************************
*************************************************
* Author:Tree *
*From :http://blog.csdn.net/lttree *
* Title : S-Nim *
*Source: hdu 1536 *
* Hint : SG *
*************************************************
*************************************************/
#include
#include
#include
using namespace std;
#define N 10001
int f[N],sg[N];
bool mex[N];
void get_sg(int t,int n)
{
int i,j;
memset(sg,0,sizeof(sg));
for(i=1;i<=n;i++)
{
memset(mex,0,sizeof(mex));
// 对于属于g(x)后继的数置1
for( j=1 ;j<=t && f[j]<=i ;j++ )
mex[sg[i-f[j]]]=1;
// 找到最小不属于该集合的数
for( j=0 ; j<=n ; j++ )
if(!mex[j])
break;
sg[i] = j;
}
}
int main()
{
int k,m,n,i,t,temp;
while( scanf("%d",&k) && k )
{
for(i=1;i<=k;++i)
scanf("%d",&f[i]);
sort(f+1,f+k+1);
get_sg(k,N);
scanf("%d",&m);
while(m--)
{
temp=0;
scanf("%d",&n);
for(i=0;i