对于\(l_1=1,r_1=1\)的情况，设\(f_{i,j,0/1,S}\)表示\(\texttt{x}\)串考虑了前\(i\)个位置，\(\texttt{y}\)串考虑了前\(j\)个位置，且最后一个位置选了\(\texttt{x}\)串还是\(\texttt{y}\)串，选的串的集合为\(S\)的方案数。转移显然。答案为\(\sum_{i=1}^n\sum_{j=1