BZOJ 1009 [HNOI2008]GT考试 ——矩阵乘法 KMP

先用KMP处理所有的转移，或者直接暴力也可以。

然后矩阵快速幂即可。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define F(i,j,k) for (ll i=j;i<=k;++i)
#define D(i,j,k) for (ll i=j;i>=k;--i)
#define ll long long
ll n,m,md;
char s[50];
ll fail[50];
struct matrix{
    ll x[25][25];
    void init(){memset(x,0,sizeof x);}
    void build1()
    {
        init();
        x[0][0]=1;
    }
    void build2()
    {
        init();
        F(i,0,m-1) F(j,0,9)
        {
            ll now=i;
            while (now&&s[now+1]-'0'!=j) now=fail[now];
            if (s[now+1]-'0'==j) now++;
            x[i][now]++;
        }
    }
    matrix operator * (matrix a) {
        matrix ret;ret.init();
        F(i,0,m-1) F(j,0,m-1)
        {
            F(k,0,m-1)
                ret.x[i][j]+=x[i][k]*a.x[k][j];
            ret.x[i][j]%=md;
        }
        return ret;
    }
    void print()
    {
        printf("---\n");
        F(i,0,m)
        {
            F(j,0,m) printf("%lld ",x[i][j]);
            printf("\n");
        }
        printf("---\n\n");
    }
    void build3()
    {
        init();
        F(i,0,m-1) x[i][i]=1;
    }
}B,A,C;
int main()
{
    scanf("%lld%lld%lld",&n,&m,&md);
    scanf("%s",s+1);
    for (int i=2,j=0;i<=m;++i)
    {
        while (j&&s[j+1]!=s[i]) j=fail[j];
        if (s[j+1]==s[i]) j++;
        fail[i]=j;
    }
    A.build1();B.build2();C.build3();
    while (n)
    {
        if (n&1) C=C*B;
        B=B*B;
        n>>=1;
    }
    A=A*C;
    int ans=0;
    F(i,0,m-1) (ans+=A.x[0][i])%=md;
    printf("%d\n",ans);
}