HDU 3306 Another kind of Fibonacci 矩阵连乘

Another kind of Fibonacci

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1496 Accepted Submission(s): 570



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



构造矩阵+矩阵连乘

#include <iostream>
#include <stdio.h>
#include <cstring>
#define Mod 10007
using namespace std;
const int MAX = 4;

typedef  struct{
         int  m[MAX][MAX];
}  Matrix;
Matrix P={1,1,0,0,
          0,0,0,0,
          0,1,0,0,
          0,0,0,0};
Matrix I={1,0,0,0,
          0,1,0,0,
          0,0,1,0,
          0,0,0,1};
Matrix matrixmul(Matrix a,Matrix b)
{
       int i,j,k;
       Matrix c;
       for (i = 0 ; i < MAX; i++)
           for (j = 0; j < MAX;j++)
             {
                 c.m[i][j] = 0;
                 for (k = 0; k < MAX; k++)
                     c.m[i][j] += (a.m[i][k]* b.m[k][j])%Mod;
                 c.m[i][j] %= Mod;
             }
       return c;
}

Matrix quickpow(long long n)
{
      Matrix m = P, b = I;
      while (n >= 1)
       {
             if (n & 1)
                b = matrixmul(b,m);
             n = n >> 1;
             m = matrixmul(m,m);
       }
       return b;
}

int main()
{
    int n,x,y,sum;
    Matrix b;
   while(cin>>n>>x>>y)
    {
      sum=0;
      x=x%Mod;
      y=y%Mod;
      P.m[1][1]=(x*x)%Mod;P.m[1][2]=(y*y)%Mod;P.m[1][3]=(2*x*y)%Mod;
      P.m[3][1]=x;P.m[3][3]=y;
      b=quickpow(n);
      for(int i=0;i<4;i++)
        sum+=b.m[0][i]%Mod;
      cout<<sum%Mod<<endl;

    }
    return 0;
}