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hdu 1536&&1944 S-Nim sg函数

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4770    Accepted Submission(s): 2058


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
 

 

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int s[101],k;int sg[10001];int n;int seg(int num){    if(sg[num]!=-1)return sg[num];    bool used[101];    memset(used,0,sizeof(used));    for(int i=0;i<k;i++){        if(num<s[i])break;        used[seg(num-s[i])]=true;    }    for(int i=0;i<101;i++)if(!used[i])return sg[num]=i;}int main(){    freopen("C:\\Users\\Administrator\\Desktop\\input.txt","w",stdout);    while(scanf("%d",&k)==1&&k){        for(int i=0;i<k;i++)scanf("%d",s+i);        memset(sg,-1,sizeof(sg));        sort(s,s+k);        int T;        scanf("%d",&T);        while(T--){            scanf("%d",&n);            int sum=0,tmp;            for(int i=0;i<n;i++){scanf("%d",&tmp);sum^=seg(tmp);}            if(sum==0)putchar(‘L‘);            else putchar(‘W‘);        }        putchar(‘\n‘);    }    return 0;}

  

hdu 1536&&1944 S-Nim sg函数