hdu 1536 SG函数模板题 千万注意游戏的组合用SG函数值异或 是SG函数的值啊
S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3077 Accepted Submission(s): 1361
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
Norgesmesterskapet 2004
Recommend
LL
题意：
首先输入K 表示一个集合的大小 之后输入集合 表示对于这对石子只能去这个集合中的元素的个数
之后输入 一个m 表示接下来对于这个集合要进行m次询问
之后m行 每行输入一个n 表示有n个堆 每堆有n1个石子 问这一行所表示的状态是赢还是输 如果赢输入W否则L
思路：
对于n堆石子 可以分成n个游戏 之后把n个游戏合起来就好了
#include
#include
#include
using namespace std;
//注意 S数组要按从小到大排序 SG函数要初始化为-1 对于每个集合只需初始化1边
//不需要每求一个数的SG就初始化一边
int SG[10100],n,m,s[102],k;//k是集合s的大小 S[i]是定义的特殊取法规则的数组
int dfs(int x)//求SG[x]模板
{
if(SG[x]!=-1) return SG[x];
bool vis[110];
memset(vis,0,sizeof(vis));
for(int i=0;i=s[i])
{
dfs(x-s[i]);
vis[SG[x-s[i]]]=1;
}
}
int e;
for(int i=0;;i++)
if(!vis[i])
{
e=i;
break;
}
return SG[x]=e;
}
int main()
{
int cas,i;
while(scanf("%d",&k)!=EOF)
{
if(!k) break;
memset(SG,-1,sizeof(SG));
for(i=0;i
下面是对SG打表的做法
#include
#include
#include
using namespace std;
const int K=101;
const int H=10001;//H是我们要打表打到的最大值
int k,m,l,h,s[K],sg[H],mex[K];///k是集合元素的个数 s[]是集合 mex大小大约和集合大小差不多
///注意s的排序
void sprague_grundy()
{
int i,j;
sg[0]=0;
for (i=1;i=s[j]){
mex[sg[i-s[j]]]=1;
j++;
}
j=0;
while (mex[j]) j++;
sg[i]=j;
}
}
int main(){
int tmp,i,j;
scanf("%d",&k);
while (k!=0){
for (i=1;i<=k;i++)
scanf("%d",&s[i]);
sort(s+1,s+k+1); //这个不能少
sprague_grundy();
scanf("%d",&m);
for (i=0;i
求SG的模板 注意理解此代码 可以在以后做题中很灵活的应用
//SG 的模板：
int mex1(int p)
{
int i,t;
bool vis[101]={0};
for(i=0;i
本题另外一种方法 用上面的模板
#include
#include
#include
using namespace std;
int f[110];
int d[20010];
bool g[110];
int mex1(int p,int k){
int i,t;
memset(g,0,sizeof(g));
for(int i = 0; i < k; i++){
t = p-f[i];
if(t < 0) break;
if(d[t] == -1) d[t] = mex1(t,k);
g[d[t]] = 1;
}
for(i = 0;;i++){
if(!g[i]) return i;
}
}
int main(){
// freopen("1.txt","r",stdin);
int n,m,k;
while(scanf("%d",&n)&&n){
for(int i = 0; i < n; i++)
scanf("%d",&f[i]);
sort(f,f+n);
memset(d,-1,sizeof(d));
d[0] = 0;
for(int i = 1; i <= 10003; i++)
d[i] = mex1(i,n);
scanf("%d",&m);
for(int i = 1; i <= m; i++){
scanf("%d",&k);
int sum = 0,v;
for(int j = 1; j <= k; j++){
scanf("%d",&v);
sum ^= d[v];
}
if(sum == 0) printf("L");
else printf("W");
}
printf("\n");
}
return 0;
}