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): 419    Accepted Submission(s): 148


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
 

 

Recommend
zhuyuanchen520
 

 http://www.cnblogs.com/liuxueyang/archive/2013/08/20/3270893.html

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int mod=1000000007;

struct Matrix{
    long long arr[5][5];
};

Matrix init,unit;
long long n,a0,ax,ay,b0,bx,by;  //注意用long long,否则会溢出

Matrix Mul(Matrix a,Matrix b){
    Matrix c;
    for(int i=0;i<5;i++)
        for(int j=0;j<5;j++){
            c.arr[i][j]=0;
            for(int k=0;k<5;k++)
                c.arr[i][j]=(c.arr[i][j]+a.arr[i][k]*b.arr[k][j]%mod)%mod;
            c.arr[i][j]%=mod;
        }
    return c;
}

Matrix Pow(Matrix a,Matrix b,long long k){
    while(k){
        if(k&1){
            b=Mul(b,a);
        }
        a=Mul(a,a);
        k>>=1;
    }
    return b;
}

void Init(){
    for(int i=0;i<5;i++)
        for(int j=0;j<5;j++){
            init.arr[i][j]=0;
            unit.arr[i][j]=0;
        }
    unit.arr[0][0]=1,   unit.arr[0][1]=a0%mod,    unit.arr[0][2]=b0%mod,    unit.arr[0][3]=a0*b0%mod,
    unit.arr[0][4]=a0*b0%mod;

    init.arr[0][0]=1,   init.arr[0][1]=ay%mod,    init.arr[0][2]=by%mod,    init.arr[0][3]=ay*by%mod,
    init.arr[0][4]=ay*by%mod,   init.arr[1][1]=ax%mod,  init.arr[1][3]=ax*by%mod,   init.arr[1][4]=ax*by%mod,
    init.arr[2][2]=bx%mod,  init.arr[2][3]=ay*bx%mod,   init.arr[2][4]=ay*bx%mod,   init.arr[3][3]=ax*bx%mod,
    init.arr[3][4]=ax*bx%mod,   init.arr[4][4]=1;
}

int main(){

    //freopen("input.txt","r",stdin);

    while(cin>>n){
        //scanf("%d%d%d%d%d%d",&a0,&ax,&ay,&b0,&bx,&by);
        cin>>a0>>ax>>ay>>b0>>bx>>by;
        if(n==0){
            puts("0");
            continue;
        }
        Init();
        Matrix ans=Pow(init,unit,n-1);
        cout<<ans.arr[0][4]<<endl;
    }
    return 0;
}

 

posted @ 2013-08-21 08:32  Jack Ge  阅读(566)  评论(0编辑  收藏  举报