hdu 4686 Arc of Dream（矩阵快速幂）

Arc of Dream

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1542    Accepted Submission(s): 517

Problem Description

An Arc of Dream is a curve defined by following function:

where
a₀ = A0
a_i = a_i-1*AX+AY
b₀ = B0
b_i = b_i-1*BX+BY
What is the value of AoD(N) modulo 1,000,000,007?

 

Input

There are multiple test cases. Process to the End of File.
Each test case contains 7 nonnegative integers as follows:
N
A0 AX AY
B0 BX BY
N is no more than 10¹⁸, and all the other integers are no more than 2×10⁹.

 

Output

For each test case, output AoD(N) modulo 1,000,000,007.

 

Sample Input

1
1 2 3
4 5 6
2
1 2 3
4 5 6
3
1 2 3
4 5 6

 

Sample Output

4
134
1902


构造矩阵，上图：




AC代码：
#include <iostream>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <queue>
#include <stack>
#include <ctime>
#include <vector>
#include <algorithm>
#define ll __int64
#define L(rt) (rt<<1)
#define R(rt)  (rt<<1|1)

using namespace std;

const int INF = 2000000000;
const int maxn = 7;
const int mod = 1e9+7;

struct Mat
{
    ll mat[maxn][maxn];
    void zero()
    {
        memset(mat, 0, sizeof(mat));
    }
    void unit()
    {
        zero();
        for(int i = 0; i < maxn; i++) mat[i][i] = 1;
    }
} A, T;
ll n, A0, B0, AX, BX, AY, BY;
Mat operator * (const Mat &a, const Mat &b)
{
    Mat tmp;
    for(int i = 0; i < maxn; i++)
        for(int j = 0; j < maxn; j++)
        {
            ll sum = 0;
            for(int k = 0; k < maxn; k++)
                    sum += a.mat[i][k] * b.mat[k][j] % mod;
            tmp.mat[i][j] = sum % mod;
        }
    return tmp;
}
Mat operator ^ (Mat x, ll n)
{
    Mat tmp;
    tmp.unit();
    while(n)
    {
        if(n & 1) tmp = tmp * x;
        x = x * x;
        n >>= 1;
    }
    return tmp;
}
void init(){
    A.zero();
    A.mat[0][0] = A0 * B0 % mod, A.mat[1][0] = A0 % mod, A.mat[2][0] = B0 % mod;
    A.mat[3][0] = AY * BY % mod, A.mat[4][0] = AY % mod, A.mat[5][0] = BY % mod, A.mat[6][0] = 0;
    T.zero();
    T.mat[0][0] = AX * BX % mod, T.mat[0][1] = AX * BY % mod, T.mat[0][2] = BX * AY % mod;
    T.mat[1][1] = AX % mod, T.mat[2][2] = BX % mod;
    for(int i = 0; i < 3; i++) T.mat[i][i + 3] = 1;
    for(int i = 3; i < maxn; i++) T.mat[i][i] = 1;
    T.mat[6][0] = 1;
}
int main()
{
    while(cin>>n)
    {
        cin>>A0>>AX>>AY;
        cin>>B0>>BX>>BY;
        if(!n)
        {
            cout<<0<<endl;
            continue;
        }
        init();
        Mat ans = (T ^ n) * A;

        cout<<ans.mat[6][0]<<endl;
    }
    return 0;
}