hdu 4986 Arc of Dream 矩阵快速幂求递推式

最新推荐文章于 2020-07-01 18:38:39 发布

HuanTongH

最新推荐文章于 2020-07-01 18:38:39 发布

阅读量569

点赞数

CC 4.0 BY-SA版权

分类专栏：矩阵动态规划文章标签：快速幂递推式

本文链接：https://blog.youkuaiyun.com/HTT_H/article/details/44840769

动态规划同时被 2 个专栏收录

15 篇文章

订阅专栏

矩阵

2 篇文章

订阅专栏

本文介绍了一种通过矩阵快速幂方法解决弧梦曲线递推问题的算法。该算法利用矩阵运算将问题复杂度降低到O(5*5*5*logn)，适用于求解大规模递推问题。文章提供了完整的C++代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Arc of Dream

Time Limit: 2000/2000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2534 Accepted Submission(s): 777

Problem Description

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

where
a ₀ = A0
a _i = a _i-1*AX+AY
b ₀ = 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 10 ¹⁸, and all the other integers are no more than 2×10 ⁹.

Output

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

Sample Input

Sample Output

Author

Zejun Wu (watashi)

Source

2013 Multi-University Training Contest 9

Recommend

zhuyuanchen520 | We have carefully selected several similar problems for you: 5197 5196 5195 5194 5193

先将式子转化为递推式, Sn = Sn-1+an-1*bn-1，其中an*bn也是递推式an*bn=ax*bx*an-1*bn-1+ax*by*an-1+ay*bx*bn-1+ay*by。

递推式可以转为矩阵快速幂O(n^3logk)复杂度，n*n的矩阵的k次幂

[sn an-1*bn-1 an-1 bn-1 1] = [ sn-1 an-2*an-2 an-2 bn-2 1]* [ [1 ax*bx ax*by ay*bx ay*by] [0 ax*bx ax*by ay*bx ay*by] [0 0 ax 0 ay] [0 0 0 bx by] [0 0 0 0 1]] //共5列

前两个矩阵为1*5,最后个矩阵为5*5.根据递推公式很容易列出第一个矩阵每一项对应的列的系数是什么，从而得到最后一个矩阵。

复杂度是O(5*5*5*logn)

#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <algorithm>

using namespace std;

#define M 1000000007

typedef unsigned long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

mat mul(mat &a, mat &b)
{
    mat c(a.size(), vec(b[0].size()));
    for(int i = 0 ;i < a.size(); i++)
        for(int j = 0; j < b.size(); j++)
        for(int k = 0; k < b[0].size(); k++)
          c[i][k] = (c[i][k]+a[i][j]*b[j][k])%M;

    return c;
}

mat power(mat a, ll n)
{
    mat  b(a.size(), vec(a.size()));
    for(int i = 0; i < a.size(); i++)
        b[i][i] = 1;
    while(n > 0){
        if(n&1) b = mul(b,a);
        a = mul(a, a);
        n >>= 1;
    }
    return b;
}

int main()
{

    ll n;
    ll a0,ax,ay,b0,bx,by;
    while(cin >> n){
        cin >> a0 >> ax>> ay >> b0 >> bx >> by;
        if(!n){
            cout << 0 <<endl;
            continue;
        }

        a0%=M,b0%=M,ax%=M,ay%=M,bx%=M,by%=M;
        mat a(5, vec(5));
        a[0][0] = 1;
        a[1][0] = ax*bx%M,a[1][1] = ax*bx%M;
        a[2][0] = ax*by%M, a[2][1]=ax*by%M,a[2][2] = ax;
        a[3][0] = ay*bx%M, a[3][1]=ay*bx%M,a[3][3] = bx;
        a[4][0] = by*ay%M, a[4][1] = by*ay%M, a[4][2] = ay, a[4][3] = by, a[4][4] = 1;


        a = power(a, n-1);
        mat fir(1, vec(5));
        fir[0][0] = a0*b0%M, fir[0][1] = a0*b0%M, fir[0][2] = a0, fir[0][3] = b0, fir[0][4] = 1;
        mat ans = mul(fir, a);
        cout << ans[0][0] << endl;
    }
    return 0;
}