hdu 4990 Reading comprehension

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4990

思路：题目难点就是找矩阵。。。。。

code:

#include<cstdio>
#include<iostream>
#include<cstring>

using namespace std;

typedef long long LL;

struct Matrix
{
    LL x[5][5];
    friend Matrix operator*(Matrix &a,Matrix &b);
    friend Matrix operator*(Matrix a,LL k);
};

int n,mod;

Matrix operator*(Matrix &a,Matrix &b)
{
    Matrix c;
    int i,j,k;
    for(i=1;i<=3;i++)
    {
        for(j=1;j<=3;j++)
        {
            c.x[i][j]=0;
            for(k=1;k<=3;k++)
            {
                c.x[i][j]=(c.x[i][j]+(a.x[i][k]*b.x[k][j])%mod)%mod;
            }
        }
    }
    return c;
}

Matrix operator^(Matrix a,LL k)
{
    Matrix unit;
    memset(unit.x,0,sizeof(unit.x));
    for(int i=0;i<=3;i++)
    {
        unit.x[i][i]=1;
    }
    while(k>0)
    {
        if(k&1) unit=unit*a;
        a=a*a;
        k=k/2;
        //printf("AAAAA\n");
    }
    return unit;
}

Matrix A,B;

int main()
{
    while(scanf("%d%d",&n,&mod)==2)
    {
        memset(A.x,0,sizeof(A.x));
        A.x[1][2]=2,A.x[2][1]=1,A.x[2][2]=1;
        A.x[3][2]=A.x[3][3]=1;
        B.x[1][1]=1%mod;
        B.x[1][2]=2%mod;
        B.x[1][3]=1;
        if(n==1)
            printf("%lld\n",B.x[1][1]);
        else if(n==2)
        {
            printf("%lld\n",B.x[1][2]);
        }
        else
        {
            A=A^(n-2);
            B=B*A;
            printf("%lld\n",B.x[1][2]%mod);
        }
    }
    return 0;
}