yysys

构造矩阵，进行矩阵快速幂即可。。。 #include #include #include #include #include #include #include #include #include #include #include #include #include #define maxn 55 #define maxm 2000

推出公式，然后用juzh

先构建AC自动机，然后通过ACzidongji

暴力就行了。。。 #include #include #include #include #include #include #include #include #include #include #include #include #include #define maxn 20005 #define maxm 4000005 #define eps

附上CF的官方题解。。。 It's easy to realize that , where dp[i] is number of vertices, which are situated on a distance i from the root, and cnt[j] is number of children, which are situated on a distance

对每一列状态压缩，然后矩阵加速即可。。。 #include using namespace std; typedef long long LL; const int mod = 1e9+7; const int maxn = 130; LL mat[maxn][maxn]; LL mid[maxn][maxn]; LL res[maxn][maxn]; int n, m; void ca

建出转移矩阵...然后做矩阵快速幂就行了.... #include using namespace std; typedef long long LL; const int maxn = 105; const int INF = 0x3f3f3f3f; int mat[maxn][maxn]; int res[maxn][maxn]; int mid[maxn][maxn]; int a[