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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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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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// 本文件是轮廓线dp的版本，轮廓线dp的讲解在每周直播课 :
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// 2022年9月第4周的课，看了就能懂
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// 这是很难的一类题型，不做要求，大厂几乎不考
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public class Code03_PavingTile2 {
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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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// 轮廓线dp的版本
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// 看课，上面说了看哪的
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public static int MAXN = 12;
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public static int MAXM = (1 << MAXN);
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public static long[][][] dp = new long[MAXN][MAXN][MAXM];
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public static long ways(int n, int m) {
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for (int i = 0; i <= n; i++) {
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for (int j = 0; j <= m; j++) {
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for (int k = 0; k <= (1 << m); k++) {
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dp[i][j][k] = -1;
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}
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}
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}
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return process(0, 0, 0, n, m);
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}
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public static long process(int r, int c, int s, int n, int m) {
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if (r == n) {
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return s == 0 ? 1 : 0;
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}
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if (c == m) {
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return process(r + 1, 0, s, n, m);
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}
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if (dp[r][c][s] != -1) {
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return dp[r][c][s];
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}
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long ans;
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int cur = (s & (1 << c)) == 0 ? 0 : 1;
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if (cur == 1) {
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ans = process(r, c + 1, s ^ (1 << c), n, m);
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} else {
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long p1 = process(r, c + 1, s | (1 << c), n, m);
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long next = c + 1 < m && (s & (1 << (c + 1))) == 0 ? 0 : 1;
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long p2 = 0;
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if (next == 0) {
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p2 = process(r, c + 2, s, n, m);
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}
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ans = p1 + p2;
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}
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dp[r][c][s] = ans;
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return ans;
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}
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}
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