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package class43;
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// 找到了贴瓷砖问题在线测试
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// 测试链接 : http://poj.org/problem?id=2411
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// 注册一个北京大学评测平台的号
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// 请同学们务必参考如下代码中关于输入、输出的处理
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// 这是输入输出处理效率很高的写法
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// 提交以下的code，提交时请把类名改成"Main"
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// 本文件是状态压缩的动态规划版本，也就是课上讲的版本
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import java.io.BufferedReader;
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import java.io.IOException;
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import java.io.InputStreamReader;
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import java.io.OutputStreamWriter;
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import java.io.PrintWriter;
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import java.io.StreamTokenizer;
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public class Code03_PavingTile1 {
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public static void main(String[] args) throws IOException {
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BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
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StreamTokenizer in = new StreamTokenizer(br);
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PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
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while (in.nextToken() != StreamTokenizer.TT_EOF) {
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int n = (int) in.nval;
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in.nextToken();
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int m = (int) in.nval;
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if (n != 0 || m != 0) {
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long ans = ways(n, m);
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out.println(ans);
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out.flush();
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}
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}
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}
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// 状态压缩动态规划，最后一个版本
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// 其实其他版本也能通过
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public static long ways(int N, int M) {
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if (N < 1 || M < 1 || ((N * M) & 1) != 0) {
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return 0;
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}
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if (N == 1 || M == 1) {
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return 1;
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}
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int big = N > M ? N : M;
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int small = big == N ? M : N;
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int sn = 1 << small;
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int limit = sn - 1;
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long[] dp = new long[sn];
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dp[limit] = 1;
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long[] cur = new long[sn];
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for (int level = 0; level < big; level++) {
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for (int status = 0; status < sn; status++) {
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if (dp[status] != 0) {
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int op = (~status) & limit;
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dfs(dp[status], op, 0, small - 1, cur);
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}
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}
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for (int i = 0; i < sn; i++) {
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dp[i] = 0;
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}
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long[] tmp = dp;
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dp = cur;
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cur = tmp;
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}
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return dp[limit];
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}
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public static void dfs(long way, int op, int index, int end, long[] cur) {
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if (index == end) {
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cur[op] += way;
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} else {
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dfs(way, op, index + 1, end, cur);
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if (((3 << index) & op) == 0) {
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dfs(way, op | (3 << index), index + 1, end, cur);
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}
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}
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}
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}
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