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 cba2 abc bca10 abc bca13 abejkcfghid jkebfghicda

样例输出：

4145207352860

经典代码：

 

#include <stdio.h>

#include <string.h>

char

pre[30], post[30];

int

m;

int

find(

char

* p,

char

x) {

    

int

i=0;

    

while

(p[i]!=x) i++;

    

return

i;

}

typedef

long

long

ll;

ll C(

int

a,

int

b) {

    

ll u = 1;

    

ll d = 1;

    

while

(b) {

    

u *= a--;

    

d *= b--;

    

}

    

return

u/d;

}

ll test(

char

* p,

char

* q,

int

n) {

    

if

(n==0)

return

1;

    

ll f = 1;

    

int

c = 0;

    

int

i;

    

while

(n) {

    

c++;

    

i = find(q,*p);

    

f *= test(p+1,q,i);

    

p += i+1;

    

q += i+1;

    

n -= i+1;

    

}

    

return

f * C(m,c);

}

int

main() {

    

while

(

scanf

(

"%d%s%s"

,&m,pre,post)==3) {

    

printf

(

"%lld\n"

,test(pre+1,post,

strlen

(pre)-1));

    

}  

    

return

0;

}

 

百度文库的解释:http://wenku.baidu.com/link?url=ZddYeW-pYEgst83coqElNsI-aHY_JwyuHwsKBHkrxPNWxYCMCn0ltDqq7K-IGdZkr48WdgG4chrIkS1h5cWUnhzPQbJBL3a4N_OLUffbe4i