HDU 3306 Another kind of Fibonacci

Another kind of Fibonacci

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2258    Accepted Submission(s): 900

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

 

Recommend

wxl

 

 

 

已知下列初始条件和递推式，求S(N)。

 

由于S(N)计算需要用到A(N)^2，计算A(N)^2，发现需要A(N-1)*A(N-2)的值，

而A(N-1)*A(N-2)可以由A(N-2)、A(N-3)运算得到。

 

又有，

 

将以上式子列成矩阵

 

得到下面含幂次式子，以便使用快速幂优化计算过程，

 

使用矩阵乘法和矩阵快速幂实现程序，注意取模。

 

 

#include <iostream>
#include <algorithm>
#include <map>
#include <vector>
#include <functional>
#include <string>
#include <cstring>
#include <queue>
#include <set>
#include <cmath>
#include <cstdio>
using namespace std;
#define IOS ios_base::sync_with_stdio(false)
const int INF=0x3f3f3f3f;

const int maxn=4;
const int modnum=10007;
int N,X,Y,T;
typedef struct matrix{
    int v[maxn][maxn];
    void init(){memset(v,0,sizeof(v));}
}M;
M mul_mod(const M &a,const M &b,int L,int m,int n)
{
    M c; c.init();
    for(int i=0;i<L;i++){
        for(int j=0;j<n;j++){
            for(int k=0;k<m;k++)//注意j，k范围
                c.v[i][j]+=a.v[i][k]*b.v[k][j];
            c.v[i][j]%=modnum;
        }
    }
    return c;
}
M power(M x,int L,int p)
{
    M tmp; tmp.init();
    for(int i=0;i<L;i++)
        tmp.v[i][i]=1;
    while(p){
        if(p&1) tmp=mul_mod(x,tmp,L,L,L);
        x=mul_mod(x,x,L,L,L);
        p>>=1;
    }
    return tmp;
}

int main()
{
    while(scanf("%d%d%d",&N,&X,&Y)==3){
        M b,a;
        X%=modnum; Y%=modnum;
        b.init();
        b.v[0][0]=2;
        b.v[1][0]=b.v[2][0]=b.v[3][0]=1;
        a.init();
        a.v[0][0]=a.v[2][1]=1;
        a.v[0][1]=a.v[1][1]=X*X%modnum;
        a.v[0][2]=a.v[1][2]=Y*Y%modnum;
        a.v[0][3]=a.v[1][3]=2*X*Y%modnum;
        a.v[3][1]=X;
        a.v[3][3]=Y;
        a=power(a,4,N-1);
        a=mul_mod(a,b,4,4,1);
        printf("%d\n",a.v[0][0]);
    }
    return 0;
}