<span size="+0"><strong><span style="font-family:Arial;font-size:12px;color:green;FONT-WEIGHT: bold">Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4479 Accepted Submission(s): 1941 </span></strong></span>
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.
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
思路：尼姆博弈的变形，套用了sg函数
代码如下：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int k,a[110],f[10010];
int mex(int p)//sg函数
{
int i,t;
bool g[110]={0};
for(i=0;i<k;i++)
{
t=p-a[i];
if(t<0)
break;
if(f[t]==-1)
f[t]=mex(t);
g[f[t]]=1;
}
for(i=0;;i++)
if(!g[i])
return i;
}
int main()
{
int n,i,m,j,t,s;
while(~scanf("%d",&k),k)
{
for(i=0;i<k;i++)
scanf("%d",&a[i]);
sort(a,a+k);
memset(f,-1,sizeof(f));
f[0]=0;
scanf("%d",&n);
while(n--)
{
scanf("%d",&m);
s=0;
while(m--)
{
scanf("%d",&t);
if(f[t]==-1)
f[t]=mex(t);
s^=f[t];//尼姆博弈，各个堆的数量的异或和为零的话，先手输
}
if(s==0)
printf("L");
else
printf("W");//题上的没有换行，注意细节 ，此处没有换行符
}
printf("\n");
}
return 0;
}原文地址：http://blog.csdn.net/ice_alone/article/details/38586707
