- 题目描述：
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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：
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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：
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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：
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2 abc cba 2 abc bca 10 abc bca 13 abejkcfghid jkebfghicda
- 样例输出：
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4 1 45 207352860
- 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
",test(pre+1,post,strlen(pre)-1)); } return 0;}