Neal is very curious about combinatorial problems, and now here comes a problem about words. Knowingthat Ray has a photographic memory and this may not trouble him, Neal gives it to Jiejie.Since Jiejie can’t remember numbers clearly, he just uses sticks to help himself. Allowing for Jiejie’sonly 20071027 sticks, he can only record the remainders of the numbers divided by total amount ofsticks.The problem is as follows: a word needs to be divided into small pieces in such a way that eachpiece is from some given set of words. Given a word and the set of words, Jiejie should calculate thenumber of ways the given word can be divided, using the words in the set.
INPUT
The input file contains multiple test cases. For each test case: the first line contains the given wordwhose length is no more than 300 000.
The second line contains an integer S, 1 ≤ S ≤ 4000.
Each of the following S lines contains one word from the set. Each word will be at most 100characters long. There will be no two identical words and all letters in the words will be lowercase.
There is a blank line between consecutive test cases.You should proceed to the end of file.You should proceed to the end of file.
OUTPUT
For each test case, output the number, as described above, from the task description modulo 20071027.
Sample Input
abcd
4
a
b
cd
ab
Sample Output
Case 1: 2
这是一道很好的题目，（竟然是用到trie和dp~~）.先把单词做成trie(字典树的形式),然后用动态规划，用dp[i]表示文本串第i个字符后面（既s[i....n]）所形成的做多数量，然后推出转移方程dp[i]=dp[i]+dp[i+len[j]];
代码如下：
#include<iostream>
#include<cstring>
#include<vector>
#include<cstdio>
using namespace std;
const int maxnode = 4000 * 100 + 10;
const int sigma_size = 26;
const int maxl = 300000 + 10;
const int maxw = 4000 + 10;
const int maxwl = 100 + 10;
const int MOD = 20071027;
int d[maxl], len[maxw], S;
char text[maxl], word[maxwl];
struct Trie {
int ch[maxnode][sigma_size];
int val[maxnode];
int sz;
void clear() { sz = 1; memset(ch[0], 0, sizeof(ch[0])); }
int idx(char c) { return c - 'a'; }
void insert(const char *s, int v)
{
int u = 0, n = strlen(s);
for (int i = 0; i < n; i++)
{
int c = idx(s[i]);
if (!ch[u][c])
{
memset(ch[sz], 0, sizeof(ch[sz]));
val[sz] = 0;
ch[u][c] = sz++;
}
u = ch[u][c];
}
val[u] = v;
}
void find(const char *s, int len, vector<int>& ans) {
int u = 0;
for (int i = 0; i < len; i++)
{
int c = idx(s[i]);
if (!ch[u][c]) break;
u = ch[u][c];
if (val[u]) ans.push_back(val[u]);
}
}
} trie;
int main()
{
int kase = 1;
while (~scanf("%s%d",text,&S))
{
trie.clear();
for (int i = 1; i <= S; i++)
{
scanf("%s", word);
len[i] = strlen(word);
trie.insert(word, i);
}
memset(d, 0, sizeof(d));
int L = strlen(text);
d[L] = 1;
for (int i = L - 1; i >= 0; i--)
{
vector<int> p;
trie.find(text + i, L - i, p);
for (int j = 0; j < p.size(); j++)
d[i] = (d[i] + d[i + len[p[j]]]) % MOD;
}
printf("Case %d: %d
", kase++, d[0]);
}
return 0;
}