The alphabet of Freeland consists of exactly N letters. Each sentence of Freeland language (also known as Freish) consists of exactly M letters without word breaks. So, there exist exactly N^M different Freish sentences.
But after recent election of Mr. Grass Jr. as Freeland president some words offending him were declared unprintable and all sentences containing at least one of them were forbidden. The sentence S contains a word W if W is a substring of S i.e. exists such k >= 1 that S[k] = W[1], S[k+1] = W[2], …,S[k+len(W)-1] = W[len(W)], where k+len(W)-1 <= M and len(W) denotes length of W. Everyone who uses a forbidden sentence is to be put to jail for 10 years.
Find out how many different sentences can be used now by freelanders without risk to be put to jail for using it.
Input
The first line of the input file contains three integer numbers: N – the number of letters in Freish alphabet, M – the length of all Freish sentences and P – the number of forbidden words (1 <= N <= 50, 1 <= M <= 50, 0 <= P <= 10).
The second line contains exactly N different characters – the letters of the Freish alphabet (all with ASCII code greater than 32).
The following P lines contain forbidden words, each not longer than min(M, 10) characters, all containing only letters of Freish alphabet.
Output
Output the only integer number – the number of different sentences freelanders can safely use.
Sample Input
2 3 1
ab
bb
Sample Output
5
建tries图后可以用矩阵乘法做的，但是毕竟乘法还是没大数加法快，就写成dp求和了，我这里是设dp[0][i]为当前所走k步时从0结点到达i结点且不走禁止结点的方法数，显然这和走k-1步的相关这里为dp[1][i]，即dp[0][i]的方法数为dp[0][u]的和((u,i)存在边时)。
#include<iostream>
#include<cstdio>
#include<string>
#include <cstring>
#include <stack>
#include <algorithm>
#include <map>
#include <queue>
using namespace std;
#define INF 0x3f3f3f3f
#define mod 10000000000
int n, m, p;
long long dp[2][105][55];
int a[105][105];
long long b[55];
int len[2][105], lenb;
int ch[105][55], last[105], val[105], f[105], sz;
int v[105];
char str[55];
map<int, int> ma;
void getch() {
scanf("%s",str);
for (int i = 0; str[i] != '\0'; ++i) {
ma[(int)str[i]] = i;
}
}
void insert() {
int u = 0;
for (int i = 0; str[i] != '\0'; ++i) {
int t = ma[(int)str[i]];
if (!ch[u][t]) {
memset(ch[sz], 0, sizeof(ch[sz]));
val[sz] = 0;
ch[u][t] = sz;
sz++;
}
u = ch[u][t];
}
val[u] = 1;
}
int bigintegeradd(long long a[], long long b[],int len) { //返回a的长度
long long p = 0;
for (int i = 0; i < len; ++i) {
a[i] = a[i] + b[i] + p;
p = 0;
if (a[i] >= mod) {
p = a[i] / mod;
a[i] %= mod;
}
}
while (p) {
a[len++] = p % mod;
p /= mod;
}
return len;
}
void bigintegermuti(long long a[], int c,int len) {
long long p = 0;
for (int i = 0; i < len; ++i) {
b[i] = a[i] * c + p;
p = 0;
if (b[i] >= mod) {
p = b[i] / mod;
b[i] %= mod;
}
}
while (p) {
b[len++] = p % mod;
p /= mod;
}
lenb = len;
}
void getfail() {
int u = 0;
queue<int> q;
for (int i = 0; i < n; ++i) {
int t = ch[u][i];
if (t) {
q.push(t);
f[t] = last[t] = 0;
}
}
while (!q.empty()) {
u = q.front();
q.pop();
for (int i = 0; i < n; ++i) {
int t = ch[u][i];
if (!t) {
ch[u][i] = ch[f[u]][i];
continue;
}
q.push(t);
int v = f[u];
while (v && !ch[v][i]) {
v = f[v];
}
f[t] = ch[v][i];
val[t] |= val[f[t]];
}
}
}
void solve() {
memset(dp, 0, sizeof(dp));
memset(len, 0, sizeof(len));
memset(a, 0, sizeof(a));
for (int i = 0; i < n; ++i) {
int t = ch[0][i];
if (!val[t]) {
dp[0][t][0]++;
len[0][t] = 1;
}
}
for (int i = 0; i < sz; ++i) {
if (val[i]) {
continue;
}
for (int j = 0; j < n; ++j) {
int t = ch[i][j];
if (!val[t]) {
a[i][t]++;
}
}
}
bool flag[105];
for (int i = 2; i <= m; ++i) {
for (int j = 0; j < sz; ++j) {
for (int k = 0; k < len[0][j]; ++k) {
dp[1][j][k] = dp[0][j][k];
}
len[1][j] = len[0][j];
}
memset(dp[0], 0, sizeof(dp[0]));
for (int j = 0; j < sz; ++j) {
if (val[j]) {
continue;
}
//memset(flag, false, sizeof(flag));
for (int k = 0; k < n; ++k) {
int t = ch[j][k];
if (!val[t]) {
//bigintegermuti(dp[1][j], a[j][t], len[1][j]);
len[0][t] = bigintegeradd(dp[0][t], dp[1][j], max(len[0][t], len[1][j]));
//flag[t] = true;
}
}
}
}
for (int i = 1; i < sz; ++i) {
len[0][0] = bigintegeradd(dp[0][0], dp[0][i], max(len[0][0], len[0][i]));
}
if (len[0][0] == 0) {
cout << 0 << endl;
return;
}
printf("%lld", dp[0][0][len[0][0] - 1]);
for (int i = len[0][0] - 2; i >= 0; --i) {
printf("%010lld", dp[0][0][i]);
}
putchar('\n');
}
int main() {
while (~scanf("%d%d%d", &n, &m, &p)) {
getch();
sz = 1;
memset(ch[0], 0, sizeof(ch[0]));
for (int i = 0; i < p; ++i) {
scanf("%s", str);
insert();
}
getfail();
solve();
ma.clear();
}
return 0;
}