【洛谷P3193】【HNOI2008】—GT考试（矩阵快速幂）

传送门

对一个串建$Ac$自动机（闲得蛋疼）
然后只需要保证不走到最后一个节点就可以了
矩阵快速幂即可

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
    static char ibuf[RLEN],*ib,*ob;
    (ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
    return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
    char ch=gc();
    int res=0,f=1;
    while(!isdigit(ch))f^=ch=='-',ch=gc();
    while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
    return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define cs const
#define bg begin
#define poly vector<int>
#define chemx(a,b) ((a)<(b)?(a)=(b):0)
#define chemn(a,b) ((a)>(b)?(a)=(b):0)
int mod;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?a-=mod:0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?a+=mod:0;}
inline int mul(int a,int b){return 1ll*a*b%mod;}
inline void Mul(int &a,int b){a=1ll*a*b%mod;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
cs int N=22;
int len;
struct mat{
	int a[N][N];
	mat(){memset(a,0,sizeof(a));}
	friend inline mat operator *(cs mat &a,cs mat &b){
		mat c;
		for(int i=0;i<=len;i++)
		for(int k=0;k<=len;k++)
		for(int j=0;j<=len;j++)
		Add(c.a[i][j],mul(a.a[i][k],b.a[k][j]));
		return c;
	}
};
namespace ac{
	int nxt[N][11],fail[N],tot;
	inline void insert(char *s){
		int p=0;
		for(int i=1,len=strlen(s+1);i<=len;i++){
			int c=s[i]-'0';
			if(!nxt[p][c])nxt[p][c]=++tot;
			p=nxt[p][c];
		}
	}
	inline void buildac(){
		queue<int> q;
		for(int i=0;i<=9;i++){
			if(nxt[0][i])q.push(nxt[0][i]);
		}
		while(!q.empty()){
			int u=q.front();q.pop();
			for(int i=0;i<=9;i++){
				int v=nxt[u][i];
				if(!v)nxt[u][i]=nxt[fail[u]][i];
				else fail[v]=nxt[fail[u]][i],q.push(v);
			}
		}
	}
	inline mat getmat(){
		mat c;
		len=tot-1;
		for(int i=0;i<tot;i++){
			for(int j=0;j<=9;j++){
				if(nxt[i][j]!=tot)Add(c.a[i][nxt[i][j]],1);
			}
		}
		return c;
	}
}
int n,m,k;
char s[N];
int main(){
	#ifdef Stargazer
	freopen("lx.cpp","r",stdin);
	#endif
	n=read(),m=read(),mod=read();
	scanf("%s",s+1);
	ac::insert(s),ac::buildac();
	mat a=ac::getmat();
	mat now;
	now.a[0][0]=1;
	for(;n;n>>=1,a=a*a)if(n&1)now=now*a;
	int res=0;
	for(int i=0;i<=len;i++)Add(res,now.a[0][i]);
	cout<<res;
}