题目描述：
We are all familiar with pre-order, in-order and post-order traversals of binary trees. A common problem in data structure classes is to find the pre-order traversal of a binary tree when given the in-order and post-order traversals. Alternatively, you
can find the post-order traversal when given the in-order and pre-order. However, in general you cannot determine the in-order traversal of a tree when given its pre-order and post-order traversals. Consider the four binary trees below:
All of these trees have the same pre-order and post-order traversals. This phenomenon is not restricted to binary trees, but holds for general m-ary trees as well.
输入：
Input will consist of multiple problem instances. Each instance will consist of a line of the form m s1 s2 indicating that the trees are m-ary trees, s1 is the pre-order traversal and s2 is the post-order traversal.All traversal strings will consist of
lowercase alphabetic characters. For all input instances, 1 <= m <= 20 and the length of s1 and s2 will be between 1 and 26 inclusive. If the length of s1 is k (which is the same as the length of s2, of course), the first k letters of the alphabet will be
used in the strings. An input line of 0 will terminate the input.
输出：
For each problem instance, you should output one line containing the number of possible trees which would result in the pre-order and post-order traversals for the instance. All output values will be within the range of a 32-bit signed integer. For each
problem instance, you are guaranteed that there is at least one tree with the given pre-order and post-order traversals.
样例输入：
2 abc cba
2 abc bca
10 abc bca
13 abejkcfghid jkebfghicda
样例输出：
4
1
45
207352860
# include<iostream> # include<string> using namespace std; int arr[21][21]; //bool flag[21]; string a1,a2; void initArr(); //int Calcom(int m,int n); long long Cal(string s1,string s2,int m) { int i,j=0,n=0; long long sum=1; int len=s1.length(); s1.erase(s1.begin()); //cout<<s1; s2=s2.substr(0,s2.length()-1); //cout<<s2; while(s1.length()>j){ //a1.erase(0,len); for(i=j;i<s1.length();i++) { if(s1[j]==s2[i]) { sum=sum*Cal(s1.substr(j,i-j+1),s2.substr(j,i-j+1),m); j=i+1; n++; break; } } // //s1=s1.substr(i+1,len-1); //s2=s2.substr(i+1,len-1); } sum=sum*arr[n][m]; return sum; } void initArr() { int i,j; arr[0][1]=arr[1][1]=1; for(i=2;i<21;i++) { arr[0][i]=1; for(j=1;j<=i;j++) { if(j==i) arr[j][i]=1; else arr[j][i]=arr[j-1][i-1]+arr[j][i-1]; } } } int main() { int m; string s1,s2; long long sum; while(cin>>m>>s1>>s2) { int len=s1.size(); if (len==0) break; initArr(); sum=Cal(s1,s2,m); cout<<sum<<endl; } return 0; }