在做这题前建议先做下poj2778
poj2778
本题跟poj2778有点类似，他是求不含模式串，长度为n的字符串的个数，本题是求至少包含一个，那么很简单，求出总数，再将不含模式串的数量减去即可。
有一些细节需要注意，首先本题的mod是2^64，所以我们把数据开ull即可。然后求总数的话也需要借助矩阵快速幂，线性递推的时间复杂度也是不可接受的。另外，本题还涉及到求和，所以状态转移矩阵中需要增加一维，第L+1列全部为1。
#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <queue>
#include <map>
#include <cstring>
#define fi first
#define se second
#define FIN freopen("in.txt","r",stdin)
#define FIO freopen("out.txt","w",stdout)
#define INF 0x3f3f3f3f
#define per(i,a,n) for(int i = a;i < n;i++)
#define rep(i,a,n) for(int i = n;i > a;i--)
#define pern(i,a,n) for(int i = a;i <= n;i++)
#define repn(i,a,n) for(int i = n;i >= a;i--)
#define fastio std::ios::sync_with_stdio(false)
#define all(a) a.begin(), a.end()
#define ll unsigned long long
#define pb push_back
#define endl "\n"
#define pii pair<int,int>
#define sc(n) scanf("%d", &n)
#define CASET int ___T; scanf("%d", &___T); for(int cs=1;cs<=___T;cs++)
template<typename T> inline void _max(T &a,const T b){
if(a<b) a = b;}
template<typename T> inline void _min(T &a,const T b){
if(a>b) a = b;}
using namespace std;
//inline ll read(){
// ll a=0;int f=0;char p=getchar();
// while(!isdigit(p)){f|=p=='-';p=getchar();}
// while(isdigit(p)){a=(a<<3)+(a<<1)+(p^48);p=getchar();}
// return f?-a:a;
//}
const int maxn = 10000*50;
//const int mod = 100000;
const int Size = 110;
struct Marix{
ll a[Size][Size];
ll n;
Marix(){
n = Size;
memset(a, 0, sizeof(a));
}
Marix(ll _n)
{
n = _n;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
a[i][j] = 0;
}
}
Marix operator * (const Marix &A){
Marix B = Marix(A.n);
for(int k = 0; k < A.n; ++k){
for(int i = 0; i < A.n; ++i){
for(int j = 0; j < A.n; ++j){
B.a[i][j] = (B.a[i][j] + (a[i][k]*A.a[k][j] ) ) ;
}
}
}
return B;
}
};
Marix mul(Marix a,Marix b){
//矩阵乘法
Marix res=Marix(a.n);
for(int i=0;i<a.n;i++){
for(int j=0;j<a.n;j++){
for(int k=0;k<a.n;k++){
ll tmp=a.a[i][k]*b.a[k][j];
res.a[i][j]=(res.a[i][j]+tmp);
}
}
}
return res;
}
Marix powMod(Marix A, ll cnt){
Marix res = Marix(A.n);
for(int i = 0; i < A.n; ++i)res.a[i][i] = 1;
while(cnt){
if(cnt&1)res = mul(res,A);
//cout << res.a[0][1] << endl;
A = mul(A,A);
cnt >>= 1;
}
return res;
}
const int maxnode = 26;
int ch[maxn][maxnode]; //字典树
int cnt[maxn]; //单词出现次数
int sz;
int fail[maxn];
//map<char,int> mp;
void init()
{
//mp['A'] = 0,mp['C']=1,mp['T']=2,mp['G']=3;
sz = 1;
memset(ch[0], 0, sizeof(ch[0]));
memset(cnt,0,sizeof(cnt));
//val[0] = 0;
cnt[0] = 0;
}
void insert(char str[], int len) //插入字符串
{
int u = 0;
per(i, 0, len)
{
int v = str[i]-'a';
if (!ch[u][v])
{
memset(ch[sz], 0, sizeof(ch[sz]));
//val[sz] = 0;
cnt[sz] = 0;
ch[u][v] = sz++;
}
u = ch[u][v];
}
cnt[u]=1;
//在这里我们可以建立一个int-string的映射，以通过节点序号得知这个点是哪个单词的结尾
}
void getfail()
{
//所有模式串已插入完成
queue<int> q;
per(i, 0, maxnode)
{
if (ch[0][i])
{
fail[ch[0][i]] = 0;
q.push(ch[0][i]);
}
}
while (!q.empty())
{
int now = q.front();
q.pop();
per(i, 0, maxnode)
{
if (ch[now][i])
{
fail[ch[now][i]] = ch[fail[now]][i];
q.push(ch[now][i]);
}
else
ch[now][i] = ch[fail[now]][i];
}
cnt[now] |= cnt[fail[now]];
}
}
Marix getmatrix()
{
Marix ma = Marix(sz+1);
per(i,0,sz)
{
per(j,0,maxnode)
{
if(cnt[ch[i][j]]==1)continue;
ma.a[i][ch[i][j]]++;
}
}
for(int i=0;i<sz+1;i++){
//再多开一维，使得第id+1列全都置为1
ma.a[i][sz]=1;
}
return ma;
}
char s[20];
int main()
{
#ifndef ONLINE_JUDGE
int startTime = clock();
FIN;
#endif
//fastio;
//忘记初始化是小狗
//freopen("out.txt","w",stdout);
//ios::sync_with_stdio(false);
int n,m;
//cout << id('0');
while(~scanf("%d%d",&n,&m)){
init();
per(i,0,n)
{
scanf("%s",s);
insert(s,strlen(s));
}
getfail();
Marix mm = getmatrix();
// per(i,0,sz)
// {
// per(j,0,sz)cout << mm.a[i][j] << ' ';
// cout << endl;
// }
Marix r1 = powMod(mm,m);
ll res = 0,ans = 0;
per(i,0,sz+1)res = (res+r1.a[0][i]);
res--;
Marix r2 = Marix(2);
r2.a[0][0] = 26,r2.a[0][1] = r2.a[1][1] = 1;
Marix r3 = powMod(r2,m);
ans = r3.a[0][0]+r3.a[0][1];
ans--;
cout << ans-res << endl;
//cout << ans-1 << endl;
}
#ifndef ONLINE_JUDGE
printf("\nTime = %dms\n", clock() - startTime);
#endif
return 0;
}