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
1 #include
2 using namespace std;
3
4 const int MAXN = 100 + 5;
5 const int MAXM = 10000 + 5;
6
7 int f[MAXN];//f[0]存合法移动个数
8 int sg[MAXM];
9 bool exist[MAXN];//hash, sg不会超过合法移动个数MAXN
10
11 void getSg(int n)
12 {
13 int i, j;
14 sg[0] = 0;
15 for (i = 1; i <= n; ++i) {
16 memset(exist, false, sizeof(exist));
17 for (j = 1; j <= f[0] && f[j] <= i; ++j) {
18 exist[sg[i - f[j]]] = true;
19 }
20 for (j = 0; j < MAXN; ++j) {
21 if (!exist[j]) {
22 sg[i] = j;
23 break;
24 }
25 }
26 }
27 }
28
29
30 int main()
31 {
32 int k;//, s;
33 int m;
34 int l, hi;
35 int i, j;
36 int sum;
37
38 while (~scanf("%d", &k)) {
39 if (k == 0) {
40 break;
41 }
42 f[0] = k;
43 for (i = 1; i <= k; ++i) {
44 scanf("%d", &f[i]);
45 }
46 sort(f + 1, f + 1 + k);
47 getSg(10000);
48
49 scanf("%d", &m);
50 for (i = 0; i < m; ++i) {
51 scanf("%d", &l);
52 sum = 0;
53 for (j = 0; j < l; ++j) {
54 scanf("%d", &hi);
55 sum ^= sg[hi];
56 }
57 if (sum != 0) {
58 printf("W");
59 } else {
60 printf("L");
61 }
62 }
63 printf("\n");
64
65 }
66 return 0;
67 }
View Code
1 #include
2 using namespace std;
3
4 const int MAXN = 100 + 5;
5 const int MAXM = 10000 + 5;
6
7 int s[MAXN];
8 int sg[MAXM];
9 int n;//s中的个数
10
11 int dfsSg(int x)
12 {
13 if (sg[x] != -1) {
14 return sg[x];
15 }
16 int i;
17 bool vis[MAXN];//sg范围
18 memset(vis, false, sizeof(vis));
19 for (i = 0; i < n && s[i] <= x; ++i) {
20 dfsSg(x - s[i]);
21 vis[sg[x - s[i]]] = true;
22 }
23 for (i = 0; i <= x; ++i) {
24 if (!vis[i]) {
25 sg[x] = i;
26 break;
27 }
28 }
29 return sg[x];
30 }
31
32
33 int main()
34 {
35 int k;//, s;
36 int m;
37 int l, hi;
38 int i, j;
39 int sum;
40
41 while (~scanf("%d", &k)) {
42 if (k == 0) {
43 break;
44 }
45 n = k;
46 for (i = 0; i < k; ++i) {
47 scanf("%d", &s[i]);
48 }
49 sort(s, s + k);
50 memset(sg, -1, sizeof(sg));
51 scanf("%d", &m);
52 for (i = 0; i < m; ++i) {
53 scanf("%d", &l);
54 sum = 0;
55 for (j = 0; j < l; ++j) {
56 scanf("%d", &hi);
57 sum ^= dfsSg(hi);
58 }
59 if (sum != 0) {
60 printf("W");
61 } else {
62 printf("L");
63 }
64 }
65 printf("\n");
66
67 }
68 return 0;
69 }
View Code
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