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
sg[x]=mex(sg[y]|y为x的后继)。
#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
queue<char>q;
int op[105];
int k;
int sg[10005];
int Hash[10005];
//sg函数模板
void sg_solve(int *s,int t,int N)
{
int i,j;
memset(sg,0,sizeof(sg));
for(i=1;i<=N;i++)//递推打表过程
{
memset(Hash,0,i+1);
for(j=0;j<t;j++)
if(i - s[j] >= 0)
Hash[sg[i-s[j]]] = 1;//标记后继
以下通过标记得出mex
for(j=0;j<=N;j++)
if(!Hash[j])
break;
sg[i] = j;
}
}
/*int mex(int x){
if(sg[x] != -1) return sg[x];
bool vis[1005];
for(int i = 0;i < 1005;i++)
vis[i] = false;
for(int i = 0;i < k;i++){
int temp = x-op[i];
if(temp < 0) break;
sg[temp] = mex(temp);
vis[sg[temp]] = true;
}
for(int i = 0;;i++)
if(!vis[i]) {sg[x] = i;break;}
return sg[x];
}*///令一个模板，没看懂。
int main()
{
while(scanf("%d",&k)!=EOF&&k!=0)
{
for(int i=0;i<k;i++)
{
scanf("%d",&op[i]);
}
sg_solve(op,k,10005);
int m;
scanf("%d",&m);
while(m--)
{
int sum=0;
int x;
scanf("%d",&x);
while(x--)
{
int mm;
scanf("%d",&mm);
sum=sum^sg[mm];
}
if(sum==0)
q.push('L');
else
q.push('W');
}
while(!q.empty())
{
printf("%c",q.front());
q.pop();
}
printf("\n");
}
}