S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4004 Accepted Submission(s): 1732Problem 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
解题思路：
这个题折腾了两三天，参考了两个模板，在这之间折腾过来折腾过去，终于把用法和需要注意的地方弄清楚了，汗。注意的是： bool类型的数组比int类型的数组快，不超时与超时的区别，在sg组合博弈时，只能在（a1,a2,a3,a4）中取，要特别注意这里面的数字是否是有序的，这个特别重要，下面贴出的两个模板对应了两种情况。最后，大数据输入还得用scanf。。
模板1（该模板注意的是一定要把 f 数组中的数字从小到大排序）：
const int N=10008;//N为所有堆最多石子的数量
int f[108],sg[N];//f[]用来保存只能拿多少个，sg[]来保存SG值
bool hash[N];//mex{}
void sg_solve(int t,int N)
{
int i,j;
memset(sg,0,sizeof(sg));
for(i=1;i<=N;i++)
{
memset(hash,0,sizeof(hash));
for(j=1;j<=t&&f[j]<=i;j++)
{
hash[sg[i-f[j]]]=1;
}
for(j=0;j<=N;j++)
if(!hash[j])
break;
sg[i] = j;
}
}模板2（该模板不需要进行排序）
const int N=10008;//N为所有堆最多石子的数量
int f[108],sg[N];//f[]用来保存只能拿多少个，sg[]来保存SG值
bool hash[N];//mex{}
void sg_solve(int t,int N)
{
int i,j;
memset(sg,0,sizeof(sg));
for(i=1;i<=N;i++)
{
memset(hash,0,sizeof(hash));
for(j=1;j<=t;j++)
if(i - f[j] >= 0)
hash[sg[i-f[j]]] = 1;
for(j=0;j<=N;j++)
if(!hash[j])
break;
sg[i] = j;
}
}代码（使用第二个模板）
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
using namespace std;
const int N=10008;//N为所有堆最多石子的数量
int f[108],sg[N];//f[]用来保存只能拿多少个，sg[]来保存SG值
bool hash[N];//mex{}
void sg_solve(int t,int N)
{
int i,j;
memset(sg,0,sizeof(sg));
for(i=1;i<=N;i++)
{
memset(hash,0,sizeof(hash));
for(j=1;j<=t;j++)
if(i - f[j] >= 0)
hash[sg[i-f[j]]] = 1;
for(j=0;j<=N;j++)
if(!hash[j])
break;
sg[i] = j;
}
}
int main()
{
int k,m,l,num,i,j;
while(scanf("%d",&k),k)
{
for(i=1;i<=k;i++)
scanf("%d",&f[i]);
sg_solve(k,N);
scanf("%d",&m);
string ans="";
for( i=1;i<=m;i++)
{
int sum=0;
scanf("%d",&l);
for( j=1;j<=l;j++)
{
scanf("%d",&num);
sum^=sg[num];
}
if(sum==0)
ans+="L";
else
ans+="W";
}
cout<<ans<<endl;
}
return 0;
}