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hdu 3306 Another kind of Fibonacci

2024-07-14 11:02:47   217人阅读

Another kind of Fibonacci

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1526    Accepted Submission(s): 583


Problem Description
As we all known , the Fibonacci series : F(0) = 1, F(1) = 1, F(N) = F(N - 1) + F(N - 2) (N >= 2).Now we define another kind of Fibonacci : A(0) = 1 , A(1) = 1 , A(N) = X * A(N - 1) + Y * A(N - 2) (N >= 2).And we want to Calculate S(N) , S(N) = A(0)2 +A(1)2+……+A(n)2.

 

Input
There are several test cases.
Each test case will contain three integers , N, X , Y .
N : 2<= N <= 231 – 1
X : 2<= X <= 231– 1
Y : 2<= Y <= 231 – 1
 

Output
For each test case , output the answer of S(n).If the answer is too big , divide it by 10007 and give me the reminder.
 

Sample Input
2 1 1 
3 2 3
 

Sample Output
6
196
 

Author
wyb
 

Source
HDOJ Monthly Contest – 2010.02.06
 

题解及代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int mod=10007;
struct mat
{
    __int64 t[4][4];
    void set()
    {
        memset(t,0,sizeof(t));
    }
} a,b;

mat multiple(mat a,mat b,__int64 n,int p)
{
    int i,j,k;
    mat temp;
    temp.set();
    for(i=0; i<n; i++)
        for(j=0; j<n; j++)
        {
            if(a.t[i][j]!=0)
                for(k=0; k<n; k++)
                    temp.t[i][k]=(temp.t[i][k]+a.t[i][j]*b.t[j][k])%p;
        }
    return temp;
}

mat quick_mod(mat b,__int64 n,__int64 m,int p)
{
    mat t;
    t.set();
    for(int i=0;i<n;i++) t.t[i][i]=1;
    while(m)
    {
        if(m&1)
        {
            t=multiple(t,b,n,p);
        }
        m>>=1;
        b=multiple(b,b,n,p);
    }
    return t;
}

void init(__int64 x,__int64 y)
{
   b.set();
   b.t[0][0]=1;
   b.t[1][0]=x*x%mod;b.t[1][1]=x*x%mod;b.t[1][2]=x;b.t[1][3]=1;
   b.t[2][0]=2*x*y%mod;b.t[2][1]=2*x*y%mod;b.t[2][2]=y;
   b.t[3][0]=y*y%mod;b.t[3][1]=y*y%mod;
}
int main()
{
    __int64 n,x,y;
    while(cin>>n>>x>>y)
    {
        x=x%mod;
        y=y%mod;
        init(x,y);
        a=quick_mod(b,4,n-1,mod);
        cout<<(2*a.t[0][0]+a.t[1][0]+a.t[2][0]+a.t[3][0])%mod<<endl;
    }
    return 0;
}
/*
A为特殊斐波那契额数列：A[0]=1,A[1]=1,A[n]=x*A[n-1]+y*A[n-2];
S[n]=∑A[i]^2=A[0]^2+A[1]^2+……+A[n]^2=S[n-1]+A[n]^2;
=S[n-1]+x^2*A[n-1]^2+2*x*y*A[n-1]*A[n-2]+y^2*A[n-2]^2;
这样我们就能构造出一个4*4的初始矩阵和一个转换矩阵
初始矩阵B=|S[2] A[1]^2 A[1]*A[0] A[0]^2|
转换一次后结果T=|S[3] A[2]^2 A[2]*A[1] A[1]^2|
其中S[3]=S[2]+x*x*A[1]^2+2*x*y*A[1]*A[0]+y*y*A[0]^2;
    A[2]^2=x*x*A[1]^2+2*x*y*A[1]*A[0]+y*y*A[0]^2;
    A[2]*A[1]=x*A[1]^2+y*A[1]*A[0];
转换矩阵Z=
|1    0    0    0|
|x^2  x^2  x    1|
|2xy  2xy  y    0|
|y^2  y^2  0    0|
矩阵快速幂就可以了。

*转载请注明出处，谢谢。
*/


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