HDOJ 1536 S-NIM SG函数

传送门http://acm.hdu.edu.cn/showproblem.php?pid=1536

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8427    Accepted Submission(s): 3532

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.

 

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 2 532 5 123 2 4 74 2 3 7 125 1 2 3 4 532 5 123 2 4 74 2 3 7 120

 

Sample Output

LWWWWL

 

Source

Norgesmesterskapet 2004

 

继续练手。我写的第一版，超时。想来原因是mex函数每一层都去计算，有大量无用的工作

#include <iostream>#include<stdio.h>#include<queue>#include<algorithm>#include<vector>#include<string>#include<string.h>#include<set>#include<deque>#include<queue>#include<vector>#include<map>using namespace std;int limit[10005];int sg[10005];int mark[10005];char ans[10005];int mex(int n,int tmp){   //memset(sg,0,sizeof(sg));   for(int i=1;i<=tmp;i++)   {       memset(mark,0,sizeof(mark));       for(int j=0;j<n;j++)       {          if(i-limit[j]<0) break;          mark[sg[i-limit[j]]]++;       }       int p=0;       while(mark[p]!=0) p++;       sg[i]=p;   }    return sg[tmp];}int main(){   int n;   while(1){    cin>>n;    int ppp;    if(n==0) break;    for(int i=0;i<n;i++)    //cin>>limit[i];    scanf("%d",&limit[i]);    sort(limit,limit+n);    int num;    ppp=0;    cin>>num;      for(int i=0;i<num;i++){          int x;          cin>>x;          int sum=0;          for(int j=0;j<x;j++)            {                int tmp;                scanf("%d",&tmp);              sum^=judge(n,tmp);            }           if(sum==0) ans[ppp++]='L';           else ans[ppp++]='W';      }      for(int i=0;i<ppp;i++)        printf("%c",ans[i]);     cout<<endl;   }   return 0;}

第二份代码，AC喽·~

#include <iostream>#include<stdio.h>#include<queue>#include<algorithm>#include<vector>#include<string>#include<string.h>#include<set>#include<deque>#include<queue>#include<vector>#include<map>using namespace std;int limit[10005];int sg[10005];//char ans[10005];int mex(int n,int tmp){     if(sg[tmp]!=-1) return sg[tmp];     int mark[1005]={0};     for(int j=0;j<n;++j)       {          if(tmp-limit[j]<0) break;          if(sg[tmp-limit[j]]==-1) mex(n,tmp-limit[j]);          mark[sg[tmp-limit[j]]]++;       }       int p=0;       while(mark[p]!=0) ++p;       sg[tmp]=p;    return p;}int main(){   int n;   while(scanf("%d",&n),n){    //cin>>n;    int ppp;    if(n==0) break;    for(int i=0;i<n;i++)    //cin>>limit[i];    scanf("%d",&limit[i]);    sort(limit,limit+n);    int num;    ppp=0;    memset(sg,-1,sizeof(sg));     //这个一定要放这里！！之前一直超时就是放下面循环里了    cin>>num;      for(int i=0;i<num;i++)      {          int x;          cin>>x;          int sum=0;          for(int j=0;j<x;j++)            {                int tmp;                scanf("%d",&tmp);              int zz=mex(n,tmp);              sum=sum^zz;            }       if(sum==0) printf("L");           else printf("W");      }         printf("\n");    }  //system("pause");   return 0;}