Total Submission(s): 5856 Accepted Submission(s): 2507
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.
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
题意：给你一个集合S，再给你几个局面，每次操作某一堆只能减去S中的一个数，问当前局面是必胜局面还是必败局面。
思路：这题是sg模板题，因为终止条件是最后不能走的人输，所以当各个游戏的sg值异或和为0时为必败局面，不为0为必胜局面，我们只要处理每个游戏的sg值就行了。处理sg值的时候有两种方法，一种是每次每个局面单独求一遍，递归出它的后继游戏的sg的值，然后它的sg值就是不包含后继状态sg值的最小非负整数，还有一种是对于一个S集合求所有状态的sg值一次求出来，然后打表，那么要求的时候就可以直接用了。
代码一：递归求sg的值
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define pi acos(-1.0)
int mex[110],sg[10010],s[110],n,m;
int get_sg(int num)
{
int i,j;
if(sg[num]!=-1)return sg[num];
bool mex[110]; //mex的大小和S集合的大小差不多
memset(mex,false,sizeof(mex));
for(i=1;i<=n;i++){
if(s[i]<=num){
mex[get_sg(num-s[i])]=true;
}
else break;
}
for(i=0;;i++){
if(!mex[i]){
sg[num]=i;return i;
}
}
}
int main()
{
int i,j,l,c,num;
while(scanf("%d",&n)!=EOF && n!=0)
{
for(i=1;i<=n;i++){
scanf("%d",&s[i]);
}
sort(s+1,s+1+n); //这里要排序，因为get_sg函数中进行到s[i]>num就break了。
memset(sg,-1,sizeof(sg));
scanf("%d",&m);
while(m--)
{
scanf("%d",&l);
num=0;
for(i=1;i<=l;i++){
scanf("%d",&c);
num=num^get_sg(c);
}
if(num==0)printf("L");
else printf("W");
}
printf("
");
}
return 0;
}
代码二：先打表求出所有局面的sg函数值。
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
#define pi acos(-1.0)
int mex[110],sg[10010],s[110],n,m;
void get_sg( )
{
int i,j;
sg[0]=0;
for(i=1;i<=10000;i++){
memset(mex,0,sizeof(mex));
for(j=1;j<=n;j++){
if(s[j]<=i){
mex[sg[i-s[j] ] ]=1;
}
else break;
}
for(j=0;;j++){
if(!mex[j]){
sg[i]=j;break;
}
}
}
}
int main()
{
int i,j,l,c,num;
while(scanf("%d",&n)!=EOF && n!=0)
{
for(i=1;i<=n;i++){
scanf("%d",&s[i]);
}
sort(s+1,s+1+n);
get_sg();
scanf("%d",&m);
while(m--)
{
scanf("%d",&l);
num=0;
for(i=1;i<=l;i++){
scanf("%d",&c);
num=num^sg[c];
}
if(num==0)printf("L");
else printf("W");
}
printf("
");
}
return 0;
}