POJ1240

Pre-Post-erous!

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 2374		Accepted: 1464

Description

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: 

    a   a   a   a

   /   /     \   \

  b   b       b   b

 /     \     /     \

c       c   c       c

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. 

Input

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. 

Output

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. 

Sample Input

2 abc cba
2 abc bca
10 abc bca
13 abejkcfghid jkebfghicda
0

Sample Output

4
1
45
207352860


这个题也比较简单，写的开心

#include <iostream>
#include <cstring>
#include <stdio.h>


using namespace std;


int m,sum;
char s1[25],s2[25];


int fd(int lpre,int rpre,int lpost,int rpost)
{
    if(lpre==rpre)
        return 1;
    int cou,nr[20],nl[20],i,j,k,ve,to;
    k=1;
    i=lpre+1;
    j=lpost;
    cou=0;
    nr[0]=j-1;
    nl[0]=i-1;
    while(i<=rpre)
    {
        char tem=s1[i];
        //cout<<i<<endl;
        for(;j<=rpost-1;j++)
        {
            i++;
            if(tem==s2[j])
            {
                cou++;
                nl[k]=i-1;
                nr[k]=j;
                k++;
                j++;
                //cout<<"sdsa"<<endl;
                break;
            }
        }


    }
    ve=1;
    for(i=0;i<cou;i++)
        ve=ve*(m-i)/(i+1);


    //cout<<cou<<endl<<ve<<endl;
    to=1;
    for(i=0;i<cou;i++){
        //cout<<nl[i]+1<<"-"<<nl[i+1]<<endl;
        //cout<<nr[i]+1<<"-"<<nr[i+1]<<endl;
        to=to*fd(nl[i]+1,nl[i+1],nr[i]+1,nr[i+1]);
    }




    return to*ve;


}


int main()
{
    cin>>m>>s1>>s2;
    int lpre,rpre,lpost,rpost,len;
    while(m!=0)
    {
        len=strlen(s1);
        lpre=0;rpre=len-1;
        lpost=0;rpost=len-1;
        cout<<fd(lpre,rpre,lpost,rpost)<<endl;
        cin>>m>>s1>>s2;
    }
    return 0;
}