HDU 4686

HDU 4686

题意很简单，给出公式，求结果，将公式转化后可以用矩阵快速幂求出来，细节问题，错了很多次，不够熟练。

#include<iostream>
#include<cstring>
#include<cmath>
#include<map>
#include<algorithm>
#include<stdio.h>
#include<string>
#pragma comment(linker,"/STACK:102400000,102400000")
using namespace std;

#define LL __int64
#define M 1000000007
LL matrix[5][5];
class Matrix
{
public:
    LL c[5][5];
    Matrix()
    {
        for(LL i = 0;i < 5;i ++)
            for(LL j = 0;j < 5;j ++)
                c[i][j] = 0;
    }
    void deal()
    {
        for(LL i = 0;i < 5;i ++)
            for(LL j = 0;j < 5;j ++)
                c[i][j] = matrix[i][j];
    }
};
Matrix mul(const Matrix & t1,const Matrix & t2)
{
    LL i,j,k;
    Matrix t3;
    for(i = 0;i < 5;i ++)
    {
        for(j = 0;j < 5;j ++)
        {
            for(k = 0;k < 5;k ++)
            {
                t3.c[i][j] = (t3.c[i][j] + t1.c[i][k]*t2.c[k][j])%M;
            }
        }
    }
    return t3;
}
Matrix power(LL n)
{
    Matrix a,b;
    b.deal();
    if(n==1)
        return b;
    else if(n%2==1)
    {
        a=(power(n/2));
        return mul(mul(a,a),b);
    }
    else
    {
        a=(power(n/2));
        return mul(a,a);
    }
}
int main()
{
    LL a0,ax,ay,b0,bx,by,a,b,c,d;
    LL n;
    while(scanf("%I64d",&n) != EOF)
    {
        
        scanf("%I64d%I64d%I64d",&a0,&ax,&ay);
        scanf("%I64d%I64d%I64d",&b0,&bx,&by);
        if(n == 0)
        {
            puts("0");
            continue;
        }
        a = ax*bx%M;
        b = ax*by%M;
        c = ay*bx%M;
        d = ay*by%M;
        memset(matrix,0,sizeof(matrix));
        matrix[0][0] = a;
        matrix[0][4] = 1;
        matrix[1][0] = b;
        matrix[1][1] = ax;
        matrix[2][0] = c;
        matrix[2][2] = bx;
        matrix[3][0] = d;
        matrix[3][1] = ay;
        matrix[3][2] = by;
        matrix[3][3] = 1;
        matrix[4][4] = 1;
        Matrix ans;
        ans.c[0][0] = a0*b0%M;
        ans.c[0][1] = a0;
        ans.c[0][2] = b0;
        ans.c[0][3] = 1;
        n--;
        if(n)
        {
            ans = mul(ans,power(n));
        }
        printf("%I64d\n",(ans.c[0][0] + ans.c[0][4])%M);
    }
    return 0;
}