HDU1536 S-Nim

1|0S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7638    Accepted Submission(s): 3215

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 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

 

 

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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<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
const int N=10005;
int n,m,cas,State,a[N],f[N],SG[N];bool mex[N];
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void GetSG(int n){
    memset(SG,0,n+1<<2);
    for(int i=1;i<=n;i++){
        memset(mex,0,n+1);
        for(int j=1;f[j]<=i;j++) mex[SG[i-f[j]]]=1;
        for(int j=0;j<=n;j++) if(!mex[j]){SG[i]=j;break;}
    }
}
int main(){
    while(1){
        n=read();
        if(!n) break;
        for(int i=1;i<=n;i++) f[i]=read();
        sort(f+1,f+n+1);f[n+1]=2e9;
        GetSG(10000); 
        cas=read();
        while(cas--){
            State=0;m=read();
            for(int i=1;i<=m;i++) a[i]=read();
            for(int i=1;i<=m;i++) State^=SG[a[i]];
            putchar(State?'W':'L');
        }
        putchar('\n');
    }
    return 0;
}

 

__EOF__

本文作者：shenben
本文链接：https://www.cnblogs.com/shenben/p/6649906.html
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