HDU 4565 So Easy!

矩阵快速幂

#include<cstdio>

#include<cstring>

#include<cmath>

#include<algorithm>

using namespace std;



long long a,b,n,m;

int R,C;

struct Matrix

{

    long long A[5][5];

    Matrix operator*(Matrix b);

};



Matrix Matrix::operator*(Matrix b)

{

    Matrix c;

    int i,j,k;

    memset(c.A,0,sizeof(c.A));

    for(i=1; i<=R; i++)

        for(j=1; j<=C; j++)

            for(k=1; k<=2; k++)

                c.A[i][j]=(c.A[i][j]+(A[i][k]*b.A[k][j])%m)%m;

    return c;

}



int main()

{

    while(~scanf("%lld%lld%lld%lld",&a,&b,&n,&m))

    {

        Matrix JZ;

        JZ.A[1][1]=(2*a)%m;

        JZ.A[1][2]=(b%m-(a*a)%m+m)%m;

        JZ.A[2][1]=1;

        JZ.A[2][2]=0;

        long long B=n-2;

        Matrix c;

        c.A[1][1]=1;

        c.A[1][2]=0;

        c.A[2][1]=0;

        c.A[2][2]=1;

        R=2,C=2;

        while(B>0)

        {

            if(B%2==1) c=c*JZ,B--;

            else JZ=JZ*JZ,B=B/2;

        }

        R=1,C=2;

        Matrix CS;

        CS.A[1][1]=(((((a*a)%m+b)%m)%m)*2)%m;

        CS.A[2][1]=(2*a)%m;



        if(n==1) printf("%d\n",CS.A[2][1]);

        else if(n==2) printf("%d\n",CS.A[1][1]);

        else

        {

            CS=c*CS;

            printf("%d\n",CS.A[1][1]);

        }

    }

    return 0;

}