/*正常求解超时，然后打表通过。自己定义状态，我的解法横木块[0,0]，竖木块[1,0]，其中1表示下层。也可以横木块[0,0]，竖木块[1,2]，不过会多出一个状态，需要3进制表示。*///打表程序#include <cstdio>#include <cstring>__int64 h, w;__int64 d[11][1 << 11];__int64 check1(__int64 x)//相连的0必须为偶数个{__int64 i;__int64 z = 0;for(i = 0; i < w; ++ i){if(x & (1 << i)){if(z % 2 != 0) return 0;z = 0;}elsez ++;}if(z % 2 != 0) return 0;return 1;}__int64 check2(__int64 x, __int64 y)//判断前后两个状态x,y是否符合条件{if(x & y) return 0;__int64 tmp = x | y;return check1(tmp);}__int64 max(__int64 a, __int64 b){return a > b ? a : b;}int main(){freopen("e://data.out", "w", stdout);for(h = 1; h <= 11; ++ h)for(w = 1; w <= 11; ++ w){__int64 up = (1 << w);__int64 i, j, k;memset(d, 0, sizeof(d));for(i = 0; i < h; ++ i){for(j = 0; j < up; ++ j){if(i == 0){if(!check1(j)) continue;d[i][j] = 1;}else{for(k = 0; k < up; ++ k){if(check2(j, k)){d[i][j] += d[i - 1][k];}}}}}printf("%I64d,", d[h - 1][0]);}return 0;}//主程序#include <cstdio>__int64 A[]={0,1,0,1,0,1,0,1,0,1,0,1,2,3,5,8,13,21,34,55,89,144,0,3,0,11,0,41,0,153,0,571,0,1,5,11,36,95,281,781,2245,6336,18061,51205,0,8,0,95,0,1183,0,14824,0,185921,0,1,13,41,281,1183,6728,31529,167089,817991,4213133,21001799,0,21,0,781,0,31529,0,1292697,0,53175517,0,1,34,153,2245,14824,167089,1292697,12988816,108435745,1031151241,8940739824,0,55,0,6336,0,817991,0,108435745,0,14479521761,0,1,89,571,18061,185921,4213133,53175517,1031151241,14479521761,258584046368,3852472573499,0,144,0,51205,0,21001799,0,8940739824,0,3852472573499,0};int main(){int h, w;while(scanf("%d%d", &h, &w) != EOF){if(!h && !w) break;h --;w --;printf("%I64d\n", A[h * 11 + w]);}return 0;}
