bzoj1009: [HNOI2008]GT考试 ac自动机+矩阵快速幂

https://www.lydsy.com/JudgeOnline/problem.php?id=1009

阿申准备报名参加GT考试，准考证号为N位数X1X2....Xn(0<=Xi<=9),他不希望准考证号上出现不吉利的数字。他的不吉利数学A1A2...Am(0<=Ai<=9)有M位，不出现是指X1X2...Xn中没有恰好一段等于A1A2...Am. A1和X1可以为0

在构造好的next图上跑矩阵快速幂即可

/**************************************************************
    Problem: 1009
    User: walfy
    Language: C++
    Result: Accepted
    Time:120 ms
    Memory:1384 kb
****************************************************************/
 
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
//#define mod 1000000007
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define fio ios::sync_with_stdio(false);cin.tie(0)
 
using namespace std;
 
const double g=10.0,eps=1e-12;
const int N=200+10,maxn=200000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f;
 
ll n,m,k;
struct Node{
    ll row,col;
    ll a[30][30];
};
Node mul(Node x,Node y,ll mod)
{
    Node ans;
    ans.row=x.row,ans.col=y.col;
    memset(ans.a,0,sizeof ans.a);
    for(int i=0;i<x.row;i++)
        for(int j=0;j<x.col;j++)
           for(int k=0;k<y.col;k++)
               ans.a[i][k]=(ans.a[i][k]+x.a[i][j]*y.a[j][k]+mod)%mod;
    return ans;
}
Node quick_mul(Node x,ll n,ll mod)
{
    Node ans;
    ans.row=x.row,ans.col=x.col;
    memset(ans.a,0,sizeof ans.a);
    for(int i=0;i<ans.col;i++)ans.a[i][i]=1;
    while(n){
        if(n&1)ans=mul(ans,x,mod);
        x=mul(x,x,mod);
        n/=2;
    }
    return ans;
}
char s[N];
struct ACM{
    int root,tot;
    int Next[N*10][10],fail[N],End[N*10];
    int newnode()
    {
        memset(Next[tot],-1,sizeof Next[tot]);
        End[tot]=0;
        return tot++;
    }
    ACM()
    {
        tot=0;
        root=newnode();
    }
    void ins()
    {
        int now=root,len=strlen(s);
        for(int i=0;i<len;i++)
        {
            if(Next[now][s[i]-'0']==-1)
                Next[now][s[i]-'0']=newnode();
            now=Next[now][s[i]-'0'];
        }
        End[now]=1;
    }
    void build()
    {
        queue<int>q;
        fail[root]=root;
        for(int i=0;i<10;i++)
        {
            if(Next[root][i]==-1)Next[root][i]=root;
            else
            {
                fail[Next[root][i]]=root;
                q.push(Next[root][i]);
            }
        }
        while(!q.empty())
        {
            int now=q.front();
            q.pop();
            if(End[fail[now]])End[now]=1;
            for(int i=0;i<10;i++)
            {
                if(Next[now][i]==-1)Next[now][i]=Next[fail[now]][i];
                else
                {
                    fail[Next[now][i]]=Next[fail[now]][i];
                    q.push(Next[now][i]);
                }
            }
        }
    }
    void solve()
    {
        Node A;
        A.row=A.col=tot;
        for(int i=0;i<tot;i++)
            for(int j=0;j<10;j++)
                if(!End[Next[i][j]])
                    A.a[i][Next[i][j]]++;
        A=quick_mul(A,n,k);
        ll ans=0;
        for(int i=0;i<tot;i++)
            ans=(ans+A.a[0][i])%k;
        printf("%lld
",ans);
    }
}ac;
int main()
{
    scanf("%lld%lld%lld%s",&n,&m,&k,s);
    ac.ins();
    ac.build();
    ac.solve();
    return 0;
}
/***********************
 
***********************/