文章目錄

• 拼圖
• 題意：
• 分析:

拼圖

題意：

分析:

這是hiho1143 && hiho 1151&&1033 骨牌覆蓋 V2的拓展
搜尋確定狀態轉移方程，快速冪加速運算

LL NN,N,M;
// 注意修改maxn 的值，要不然容易T
// 注意maxn值過大，棧可能會不夠
const int maxn = 200;
struct Matrix{
  int n,m;
  Matrix(int nn = 1,int mm = 1):n(nn)
 
,m(mm){ memset(a,0,sizeof(a));};
  long long a[maxn][maxn];
};
void print(const Matrix &a)
{
 for(int i = 1;i <= a.n; ++i,cout<<endl)
  // cout<<i<<"  ";
 {
  cout<<i<<"  ";
   for(int j= 1;j <= a.m; ++j)
     // cout<<a.a[i][j]<<" ";
    printf("%2lld "
 
,a.a[i][j]);
 }
  // cout<<e
}
Matrix operator*(Matrix a,Matrix b)
{
  Matrix c(a.n,b.m);
  for(int i = 1;i <= a.n; ++i)
  {
    for(int j = 1;j <= b.m; ++j)
    {
      for(int k = 1;k <= a.m; ++k)
      {
        c.a[i][j] += a.a[i][k] * b.a[k][j];
        c.a[i][j] %= mod;
      }
    }
 

  }
//  print(c);
  return c;
}

int MM[maxn][maxn];

void dfs(int a,int now,int nxt){

    if(((now&(NN-1) )== (NN-1)))
    {
      // if(a == 4)
      //   cout<<nxt<<endl;
      // cout<<a<<endl;
      MM[a][nxt]++;
      return ;
    }
    // cout<<now<<endl;
    for(int i = 0;i < N; ++i){
      // 情況1
      // 01
        if((now>>i) & 1) continue;
        if(i != 0 && !((nxt >>(i-1)) &1))
          dfs(a,now|(1<<i),nxt|(1<<i)|(1<<(i-1)));
        if(i + 1 < N && ((now >> (i+1))&1))
          dfs(a,now|(1<<i),nxt|(1<<i)|(1<<(i+1)));
        if(i + 1 < N && !(now>>(i+1)&1)){
              dfs(a,now|(1<<i)|(1<<(i+1)),nxt|(1<<i));
              dfs(a,now|(1<<i)|(1<<(i+1)),nxt|(1<<(i+1)));
              if(i+2 < N && !(now>>(i+2)&1)){
                dfs(a,now|(1<<i)|(1<<(i+1))|(1<<(i+2)),nxt|(1<<i)|(1<<(i+1))|(1<<(i+2)));
              }
        }
        break;
    }
}
Matrix ans,B;
int main(void)
{
    scanf("%lld %lld",&M,&N);
    NN = 1<<N;
    // cout<<NN<<endl;
    for(int i = 0;i < NN; ++i){
       dfs(i,i,0);
    }
    ans.n = ans.m = B.n = B.m = NN;
    for(int i = 1;i <= NN; ++i){
      for(int j = 1;j <= NN; ++j){
        B.a[i][j] = MM[i-1][j-1];
      }
    }
    // for(int i = 1;i <= NN; ++i)
    //   printf("%2d ",i);
    // cout<<endl;
    // print(B);
    ans.a[1][1] = 1;
    while(M > 0){
      if(M & 1)
        ans = ans*B;
      B = B*B;
      M >>= 1;

    }

    cout<<ans.a[1][1]<<endl;

   return 0;
}