poj2960 S-Nim
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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:
recently learned an easy way to always be able to find the best move:
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.
- 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.
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.
- 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.
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.
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.
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
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; #define N 105 #define M 10005 int 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; }
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