算法：
在之前搜索出状态的基础上，再压缩一次状态。
View Code //by yefeng #include<iostream> using namespace std;typedef long long LL; const int mod = 9937; int mask,idx, n , m;struct Matrix{int mat[257][257];void zero(){ memset(mat,0,sizeof(mat)); } void unit(){memset(mat,0,sizeof(mat));for(int i = 0; i <= mask; i++) mat[i][i] = 1; } }A,T; Matrix operator *(const Matrix &a,const Matrix &b){Matrix tmp; tmp.zero();for(int i = 0; i < idx; i++){for(int k = 0; k < idx; k++){if( a.mat[i][k] == 0 ) continue;for(int j = 0; j < idx; j++){tmp.mat[i][j] += (a.mat[i][k]*b.mat[k][j])%mod;tmp.mat[i][j] %= mod; } } } return tmp; } Matrix operator ^(Matrix x,int k){Matrix tmp; tmp.unit();while(k){if(k&1) tmp = tmp*x;x = x*x; k >>= 1; } return tmp; } int vis[300],id[300],v[300];//检查横放状态 bool check(int x){bool flag=true;for(int i = 0; i < m; i++){if( !( x &(1<<i) ) ) continue; //如果第i位不为1,就continue if( (i == m-1) || !( x & (1<<(i+1)) ) ){ //判断第i + 1位是否为1. flag = false; break; }i++;} return flag; } void DFS(int x){if(vis[x]) return; vis[x]=1;if( id[x] == -1 ) id[x] = idx++;int c = (~x) & mask;for(int i = 0; i <= mask; i++){if( (i&c) == c && check( i&x ) ){ //i&c == c保证两层之间不能同时有0 if(id[i] == -1) id[i] = idx++;T.mat[ id[i] ][ id[x] ] += 1;if( !vis[i] ) DFS(i); } } } void init(){T.zero(); idx = 0;memset(vis,0,sizeof(vis));memset(id,0xff,sizeof(id));memset(v,0,sizeof(v));idx = 0; DFS(mask); for(int i = 0; i <= mask; i++){if( ~id[i] && T.mat[id[i]][0] )v[ id[i] ] = 1; } } void solve(int k){A = T^(k-1);LL res = 0;for(int i = 0; i < idx; i++) res = ( res+ v[i]*A.mat[i][0] ) % mod;printf("%I64d\n", res); }int main(){//freopen("F.in", "r", stdin );//freopen("1.out", "w", stdout );int a, b;while( ~scanf("%d%d",&a,&b)){n = max(a,b), m = min(a,b); if( m == 1 ){if(n&1) puts("0");else puts("1"); continue;}mask = (1<<m)-1;init();solve(n); } }
