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HDOJ 4686 Arc of Dream 矩阵快速幂

2024-07-31 06:19:37   219人阅读

矩阵快速幂:


根据关系够建矩阵 , 快速幂解决.


Arc of Dream

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2164    Accepted Submission(s): 680


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

where
a0 = A0
ai = ai-1*AX+AY
b0 = B0
bi = bi-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
 


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

using namespace std;

typedef long long int LL;

const LL mod=1000000007LL;

struct Matrix
{
    int x,y;
    LL m[6][6];
    Matrix() {x=y=5;memset(m,0,sizeof(m));}
    void one()
    {
        for(int i=0;i<5;i++) m[i][i]=1LL;
    }
    void show()
    {
        cout<<x<<" * "<<y<<endl;
        for(int i=0;i<x;i++)
        {
            for(int j=0;j<y;j++)
                cout<<m[i][j]<<",";
            cout<<endl;
        }
    }
};

Matrix Mul(Matrix& a,Matrix& b)
{
    Matrix ret;
    ret.x=a.x; ret.y=b.y;
    for(int i=0;i<a.x;i++)
    {
        for(int j=0;j<b.y;j++)
        {
            LL temp=0;
            for(int k=0;k<b.y;k++)
            {
                temp=(temp+(a.m[i][k]*b.m[k][j])%mod)%mod;
            }
            ret.m[i][j]=temp%mod;
        }
    }
    return ret;
}

Matrix quickPow(Matrix m,LL x)
{
    Matrix e;
    e.one();
    while(x)
    {
        if(x&1LL) e=Mul(e,m);
        m=Mul(m,m);
        x/=2LL;
    }
    return e;
}

LL n,A0,B0,AX,AY,BX,BY;

Matrix init_matrix()
{
    Matrix ret;
    ret.m[0][0]=1;
    ret.m[1][0]=AY; ret.m[1][1]=AX;
    ret.m[2][0]=BY; ret.m[2][2]=BX;
    ret.m[3][0]=(BY*AY)%mod; ret.m[3][1]=(AX*BY)%mod;
    ret.m[3][2]=(BX*AY)%mod; ret.m[3][3]=(AX*BX)%mod;
    ret.m[4][3]=1LL; ret.m[4][4]=1LL;
    return ret;
}

Matrix Beg()
{
    Matrix beg;
    beg.m[0][0]=1;
    beg.m[1][0]=A0;
    beg.m[2][0]=B0;
    beg.m[3][0]=A0*B0%mod;
    return beg;
}

int main()
{
    while(cin>>n)
    {
        cin>>A0>>AX>>AY>>B0>>BX>>BY;
        A0=A0%mod; AX=AX%mod; AY=AY%mod;
        B0=B0%mod; BX=BX%mod; BY=BY%mod;
        Matrix m=init_matrix();
        m=quickPow(m,n);
        Matrix beg=Beg();
        LL ans=0;
        for(int i=0;i<5;i++)
            ans=(ans+beg.m[i][0]*m.m[4][i]%mod)%mod;
        cout<<ans<<endl;
    }
    return 0;
}



HDOJ 4686 Arc of Dream 矩阵快速幂


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