hdu 1536/ hdu 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): 7751    Accepted Submission(s): 3266

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

 

 

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 <bits/stdc++.h>
using namespace std;

const int MAXN = 100 + 5;
const int MAXM = 10000 + 5;

int f[MAXN];//f[0]存合法移动个数
int sg[MAXM];
bool exist[MAXN];//hash, sg不会超过合法移动个数MAXN

void getSg(int n)
{
    int i, j;
    sg[0] = 0;
    for (i = 1; i <= n; ++i) {
        memset(exist, false, sizeof(exist));
        for (j = 1; j <= f[0] && f[j] <= i; ++j) {
            exist[sg[i - f[j]]] = true;
        }
        for (j = 0; j < MAXN; ++j) {
            if (!exist[j]) {
                sg[i] = j;
                break;
            }
        }
    }
}


int main()
{
    int k;//, s;
    int m;
    int l, hi;
    int i, j;
    int sum;

    while (~scanf("%d", &k)) {
        if (k == 0) {
            break;
        }
        f[0] = k;
        for (i = 1; i <= k; ++i) {
            scanf("%d", &f[i]);
        }
        sort(f + 1, f + 1 + k);
        getSg(10000);

        scanf("%d", &m);
        for (i = 0; i < m; ++i) {
            scanf("%d", &l);
            sum = 0;
            for (j = 0; j < l; ++j) {
                scanf("%d", &hi);
                sum ^= sg[hi];
            }
            if (sum != 0) {
                printf("W");
            } else {
                printf("L");
            }
        }
        printf("
");

    }
    return 0;
}

#include <bits/stdc++.h>
using namespace std;

const int MAXN = 100 + 5;
const int MAXM = 10000 + 5;

int s[MAXN];
int sg[MAXM];
int n;//s中的个数

int dfsSg(int x)
{
    if (sg[x] != -1) {
        return sg[x];
    }
    int i;
    bool vis[MAXN];//sg范围
    memset(vis, false, sizeof(vis));
    for (i = 0; i < n && s[i] <= x; ++i) {
        dfsSg(x - s[i]);
        vis[sg[x - s[i]]] = true;
    }
    for (i = 0; i <= x; ++i) {
        if (!vis[i]) {
            sg[x] = i;
            break;
        }
    }
    return sg[x];
}


int main()
{
    int k;//, s;
    int m;
    int l, hi;
    int i, j;
    int sum;

    while (~scanf("%d", &k)) {
        if (k == 0) {
            break;
        }
        n = k;
        for (i = 0; i < k; ++i) {
            scanf("%d", &s[i]);
        }
        sort(s, s + k);
        memset(sg, -1, sizeof(sg));
        scanf("%d", &m);
        for (i = 0; i < m; ++i) {
            scanf("%d", &l);
            sum = 0;
            for (j = 0; j < l; ++j) {
                scanf("%d", &hi);
                sum ^= dfsSg(hi);
            }
            if (sum != 0) {
                printf("W");
            } else {
                printf("L");
            }
        }
        printf("
");

    }
    return 0;
}