可以用状压dp，但是要打一下表。暴力枚举行、这一行的状态、上一行的状态，判断如果上一行的状态能转移到这一行的状态就转移。
状态定义：ans[i][S]表示i行前已经全部填满，i行已经填上的列为集合S。如果有竖着的，全部当做用这一行的去补满上一行缺的。
（貌似还是插头dp的入门题）
#include<cstdio>
#include<cstring>
typedef long long LL;
LL f[][];
LL h,w;
/*LL ans[12][2050];
LL h,w;
bool judge(LL a,LL b)
{
LL i,num=0,t1,t2;
for(i=1;i<=w;i++)
{
t1=a&1;
t2=b&1;
a>>=1;
b>>=1;
if(t1==1&&t2==1)
num++;
else if(t1==0&&t2==0)
return false;
else
{
if(num&1)
return false;
num=0;
}
}
if(num&1)
return false;
else
return true;
}
LL get(LL h,LL w)
{
LL i,j,k;
memset(ans,0,sizeof(ans));
ans[0][(1<<w)-1]=1;
for(i=1;i<=h;i++)
for(j=0;j<(1<<w);j++)
for(k=0;k<(1<<w);k++)
if(judge(k,j))
{
ans[i][j]+=ans[i-1][k];
}
return ans[h][(1<<w)-1];
}
int main()
{
for(h=1;h<=11;h++)
for(w=1;w<=11;w++)
printf("f[%lld][%lld]=%lld;\n",h,w,get(h,w));
return 0;
}*/
int main()
{
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scanf("%lld%lld",&h,&w);
while(h!=&&w!=)
{
printf("%lld\n",f[h][w]);
scanf("%lld%lld",&h,&w);
}
}