HDU1536：S-Nim（sg函数）

S-Nim

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10368    Accepted Submission(s): 4262

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1536

Descripion:

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

 

题意：

n堆石子，每堆石子都有相应的个数，然后现在每个人选择一堆从种取石子，注意这里他一开始给出了一个集合S，取的个数是集合S里面的数。最后问先手还是后手赢。

 

题解：

sg函数的一个简单应用吧，最后求sg函数的时候稍微修改一下就行了。

具体见代码吧。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
typedef long long ll;
const int N = 10005;
int n, m;
int s[N], sg[N];
short check[N];
void Get_sg() {
    memset(sg, 0, sizeof(sg));
    for(int i = 1; i < N; i++) {
        memset(check, 0, sizeof(check));
        for(int j = 1; j <= n && s[j] <= i; j++) {
            check[sg[i - s[j]]] = 1;
        }
        for(int j = 0; j < N; j++) {
            if(!check[j]) {
                sg[i] = j;
                break ;
            }
        }
    }
}
int main() {
    while(scanf("%d", &n) && n) {
        for(int i = 1; i <= n; i++)
            scanf("%d", &s[i]);
        sort(s + 1, s + n + 1);
        Get_sg();
        scanf("%d", &m);
        while(m--) {
            int k, t, x = 0;
            scanf("%d", &k);
            for(int i = 1; i <= k; i++) {
                scanf("%d", &t);
                x ^= sg[t];
            }
            if(x)
                printf("W");
            else
                printf("L");
        }
        printf("\n");
    }
    return 0;
}