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
模板题
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <set>
#include <vector>
using namespace std;
typedef long long ll;
inline int getint() {
int x=0,f=1;
char ch=getchar();
while (ch>'9' || ch<'0') {if (ch=='-') f=-1;ch=getchar();}
while (ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();}
return x*f;
}
#define re(i,ss,dd) for(int i=ss;i<dd;++i)
#define er(i,ss,dd) for(int i=ss;i>=dd,--i)
#define wl(ww) while(ww)
int n,a[10010],sg[10010];
void ssg(int x){
if(sg[x]!=-1)
return ;
bool b[10010];//找到不在集合的最小的非负数
memset(b,0,sizeof(b));
re(i,1,n+1){
int w=x-a[i];
if(w<0)
continue;
if(sg[w]==-1){
ssg(w);
}
b[sg[w]]=1;
}
re(i,0,10010){
if(b[i]==0){
sg[x]=i;
return ;
}
}
}
int main(){
int m,k;
while(scanf("%d",&n)!=EOF && n){
re(i,1,n+1) scanf("%d",&a[i]);
re(i,0,10010) sg[i]=-1;
sg[0]=0;
scanf("%d",&m);
wl(m--){
scanf("%d",&k);
int sum=0;
int x;
wl(k--){
scanf("%d",&x);
if(sg[x]==-1) ssg(x);
sum^=sg[x];
}
if(sum==0) printf("L");
else printf("W");
}
printf("\n");
}
}