HDU 4990 Reading Comprehension

Question link: http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4990

Train of Thought: the difficulty of the question is to find the matrix .....

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;}

HDU 4990 Reading Comprehension