BZOJ1009

传送门：BZOJ1009

KMP构造转移矩阵，矩阵加速DP。
以后题解都会简单一些。

/**************************************************************    Problem: 1009    User: Jerusalem    Language: C++    Result: Accepted    Time:56 ms    Memory:1492 kb****************************************************************/#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <algorithm>#include <iostream>using namespace std;int mod;struct Matrix{    int m[25][25];    Matrix(){        memset(m,0,sizeof(m));    }    Matrix C(int a,int b,int c,int d,Matrix o){        Matrix q;        for(int i=0;i<a;i++)            for(int j=0;j<b;j++)                for(int k=0;k<d;k++){                    q.m[i][k]+=((m[i][j]*o.m[j][k])%mod);                    q.m[i][k]%=mod;                }        return q;    }};char num[25];int pre[25];int n,m;int f[25];int g[25];int ans;Matrix E;Matrix M;void Make_Table(){    pre[1]=0;int k=0;    for(int i=2;i<=m;i++){        while(k&&num[k+1]!=num[i])            k=pre[k];        if(num[k+1]==num[i])            k++;        pre[i]=k;    }    for(int i=0;i<m;i++)        for(int j=0;j<=9;j++){            k=i;            while(k&&num[k+1]!=j)                k=pre[k];            if(num[k+1]==j)                M.m[i][k+1]++;            else                M.m[i][0]++;        }    for(int i=0;i<m;i++)        E.m[i][i]=1;}Matrix Power(Matrix A,int b){    if(b==0)        return E;    if(b==1)        return A;    Matrix mid=Power(A,b/2);    if(b%2==0)        return mid.C(m,m,m,m,mid);    return ((mid.C(m,m,m,m,mid)).C(m,m,m,m,A));}void Solve(){    M=Power(M,n);    for(int i=0;i<m;i++){        ans+=M.m[0][i];        ans%=mod;    }    printf("%d\n",ans);}void Readdata(){    //freopen("loli.in","r",stdin);    scanf("%d%d%d\n",&n,&m,&mod);    for(int i=1;i<=m;i++)        num[i]=getchar()-'0';}void Close(){    fclose(stdin);    fclose(stdout);}int main(){    Readdata();    Make_Table();    Solve();    Close();    return 0;}