A

Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 'toilet series' (where he had to use his toilet paper to draw on, for all of his paper was filled with squares and rectangles), he dreamt of filling a large rectangle with small rectangles of width 2 and height 1 in varying ways. 

Expert as he was in this material, he saw at a glance that he'll need a computer to calculate the number of ways to fill the large rectangle whose dimensions were integer values, as well. Help him, so that his dream won't turn into a nightmare!

Input

The input contains several test cases. Each test case is made up of two integer numbers: the height h and the width w of the large rectangle. Input is terminated by h=w=0. Otherwise, 1<=h,w<=11.

Output

For each test case, output the number of different ways the given rectangle can be filled with small rectangles of size 2 times 1. Assume the given large rectangle is oriented, i.e. count symmetrical tilings multiple times.

Sample Input

1 21 31 42 22 32 42 114 110 0

Sample Output

10123514451205

题意：

给你一个矩形的高h和宽w，问用高为1和宽为2的小矩形组成所给的矩形共有多少种方法？

思路：

用2进制的01表示不放还是放
第i行只和i-1行有关
枚举i-1行的每个状态，推出由此状态能达到的i行状态
如果i-1行的出发状态某处未放，必然要在i行放一个竖的方块，所以我对上一行状态按位取反之后的状态就是放置了竖方块的状态。
然后用搜索扫一道在i行放横着的方块的所有可能，并且把这些状态累加上i-1的出发状态的方法数，如果该方法数为0，直接continue。

代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <stack>using namespace std;int n,m;__int64 dp[15][1<<11],tem;void dfs(int i,int p,int k){    if(k>=m)    {        dp[i][p]+=tem;        return;    }    dfs(i,p,k+1);    if(k<=m-2 && !(p&1<<k) && !(p&1<<k+1))        dfs(i,p|1<<k|1<<k+1,k+2);}int main(){    ios::sync_with_stdio(false);    int i,j;    while(cin>>n>>m,n+m)    {        memset(dp,0,sizeof(dp));        tem=1;        dfs(1,0,0);        for(i=2;i<=n;i++)        {            for(j=0;j<1<<m;j++)            {                if(dp[i-1][j])                    tem = dp[i-1][j];                else                    continue;                dfs(i,~j&((1<<m)-1),0);            }        }        cout<<dp[n][(1<<m)-1]<<endl;    }    return 0;}