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poj2960 S-Nim

2024-10-04 08:22:39   209人阅读

S-Nim
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 4361 Accepted: 2296

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;#define N 105#define M 10005int s[N], sn;int sg[M];void getsg(int n){    int mk[M];    sg[0] = 0;//主要是让终止状态的sg为0     memset(mk, -1, sizeof(mk));    for(int i = 1; i < M; i++)    {        for(int j = 0; j < n && s[j] <= i; j++)            mk[sg[i-s[j]]]=i;//将所有后继的sg标记为i，然后找到后继的sg没有出现过的最小正整数                              //优化：注意这儿是标记成了i，刚开始标记成了1，这样每次需初始化mk，而标记成i就不需要了         int j = 0;        while(mk[j] == i) j++;        sg[i] = j;    }}int main(){    while(~scanf("%d", &sn), sn)    {        for(int i = 0; i < sn; i++) scanf("%d", &s[i]);        sort(s, s+sn);//排序算一个优化，求sg的时候会用到         getsg(sn);        int m;        scanf("%d", &m);        char ans[N];        for(int c = 0; c < m; c++)        {            int n, tm;            scanf("%d", &n);            int res = 0;            for(int i = 0; i < n; i++)            {                scanf("%d", &tm);                res ^= sg[tm];            }            if(res == 0) ans[c] = ‘L‘;            else ans[c] = ‘W‘;        }        ans[m]=0;        printf("%s\n", ans);    }    return 0;}

 

poj2960 S-Nim


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