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 (ie 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, eg 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
就是sg函数的运用，知识变成多组例子而已，记住一点，不能取的点就是0
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<cmath>
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define pb push_back
#define mm(a,b) memset((a),(b),sizeof(a))
#include<vector>
typedef long long ll;
typedef double db;
const ll mod=1e9+7;
using namespace std;
const double pi=acos(-1.0);
int a[105],sg[10005],d[105],heap[10005],cas;
int getsg(int n)
{
mm(d,0);
int k=0;
for(int i=0;i<cas;i++)
{
if(a[i]>n)
break;
if(n>=a[i])
{
d[k++]=sg[n-a[i]];
}
}
if(k==0) return 0;
d[k]=mod;
sort(d,d+k);
if(d[0]!=0) return 0;
for(int i=0;i<k;i++)
if(d[i+1]-d[i]>1)
return d[i]+1;
}
void first(int n)
{
mm(sg,0);
sg[0]=0;
for(int i=1;i<=10000;i++)
{
sg[i]=getsg(i);
}
}
int main()
{
int n,num;
while(1)
{
sf("%d",&cas);
if(!cas) return 0;
for(int i=0;i<cas;i++)
sf("%d",&a[i]);
sort(a,a+cas);
first(cas);
sf("%d",&n);
for(int i=0;i<n;i++)
{
sf("%d",&num);
int x=0;
for(int j=0;j<num;j++)
{
sf("%d",&heap[j]);
x^=sg[heap[j]];
}
if(x)
pf("W");
else
pf("L");
}
pf("\n");
}
}