HDOJ 1536 S-Nim

S-Nim

Time Limit: 1000MS Memory Limit: 32768KB 64bit IO Format: %I64d & %I64u

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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

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SG函数入门题。。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;int s[110],n,mm,m,sg[11000];void get_SG(){    memset(sg,0,sizeof(sg));    bool vis[11000];    for(int x=1;x<=10000;x++)    {        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)        {            if(s[i]>x) break;            vis[sg[x-s[i]]]=1;        }        for(int i=0;i<=10000;i++)        {            if(!vis[i])            {                sg[x]=i;                break;            }        }    }}int main(){while(scanf("%d",&n)!=EOF&&n){    for(int i=0;i<n;i++) scanf("%d",s+i);    sort(s,s+n);    get_SG();    scanf("%d",&mm);    while(mm--)    {        scanf("%d",&m);        int ANS=0;        while(m--)        {            int a;            scanf("%d",&a);            ANS^=sg[a];        }        if(ANS) putchar('W');        else putchar('L');    }    putchar(10);}    return 0;}

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;int s[110],n,mm,m,sg[11000];int SG_dfs(int x){    if(sg[x]!=-1) return sg[x];    bool vis[110];    memset(vis,0,sizeof(vis));    for(int i=0;i<n;i++)    {        if(x-s[i]>=0)        {            vis[SG_dfs(x-s[i])]=1;        }        else break;    }    for(int i=0;i<=10000;i++)    {        if(!vis[i])        {            sg[x]=i;            break;        }    }    return sg[x];}int main(){while(scanf("%d",&n)!=EOF&&n){    memset(sg,-1,sizeof(sg));    for(int i=0;i<n;i++) scanf("%d",s+i);    sort(s,s+n);    scanf("%d",&mm);    while(mm--)    {        scanf("%d",&m);        int ANS=0;        while(m--)        {            int a;            scanf("%d",&a);            ANS^=SG_dfs(a);        }        if(ANS) putchar('W');        else putchar('L');    }    putchar(10);}    return 0;}