HDU4686 Arc of Dream —— 矩阵快速幂

题目链接：https://vjudge.net/problem/HDU-4686

Arc of Dream

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 5506    Accepted Submission(s): 1713

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

 

Author

Zejun Wu (watashi)

 

Source

2013 Multi-University Training Contest 9

题解：

学习之处：

矩阵所要维护的，要么为变量，要么为常数1，而不是变量再乘上一个系数，或者是一个非1的常数。因为：假如变量需要乘上一个系数，那么可以在n*n矩阵中乘上。同样，如果变量需要加上一个常数，那么在对应1的位置，填上这个常数即可。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 1e6+100;

const int Size = 5;
struct MA
{
    LL mat[Size][Size];
    void init()
    {
        for(int i = 0; i<Size; i++)
        for(int j = 0; j<Size; j++)
            mat[i][j] = (i==j);
    }
};

MA mul(MA x, MA y)
{
    MA ret;
    memset(ret.mat, 0, sizeof(ret.mat));
    for(int i = 0; i<Size; i++)
    for(int j = 0; j<Size; j++)
    for(int k = 0; k<Size; k++)
        ret.mat[i][j] += 1LL*x.mat[i][k]*y.mat[k][j]%MOD, ret.mat[i][j] %= MOD;
    return ret;
}

MA qpow(MA x, LL y)
{
    MA s;
    s.init();
    while(y)
    {
        if(y&1) s = mul(s, x);
        x = mul(x, x);
        y >>= 1;
    }
    return s;
}

int main()
{
    LL n, a0, ax, ay, b0, bx, by;
    while(scanf("%lld",&n)!=EOF)
    {
        scanf("%lld%lld%lld", &a0,&ax,&ay);
        scanf("%lld%lld%lld", &b0,&bx,&by);
        a0 %= MOD; ax %= MOD; ay %= MOD;
        b0 %= MOD; bx %= MOD; by %= MOD;

        if(n==0)
        {
            printf("%lld
", 0LL);
            continue;
        }

        MA s;
        memset(s.mat, 0, sizeof(s.mat));
        s.mat[0][0] = 1;
        s.mat[0][1] = s.mat[1][1] = 1LL*ax*bx%MOD;
        s.mat[0][2] = s.mat[1][2] = 1LL*ax*by%MOD;
        s.mat[0][3] = s.mat[1][3] = 1LL*ay*bx%MOD;
        s.mat[0][4] = s.mat[1][4] = 1LL*ay*by%MOD;
        s.mat[2][2] = ax; s.mat[2][4] = ay;
        s.mat[3][3] = bx; s.mat[3][4] = by;
        s.mat[4][4] = 1;

        LL f0, s0;
        s0 = f0 = 1LL*a0*b0%MOD;
        s = qpow(s, n-1);
        LL ans = 0;
        ans += (1LL*s0*s.mat[0][0]%MOD+1LL*f0*s.mat[0][1]%MOD)%MOD, ans %= MOD;
        ans += (1LL*a0*s.mat[0][2]%MOD+1LL*b0*s.mat[0][3]%MOD)%MOD, ans %= MOD;
        ans += 1LL*s.mat[0][4]%MOD, ans %= MOD;
        printf("%lld
", ans);
    }
}