HDU 3306 Another kind of Fibonacci 矩阵连乘

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本文探讨了一种使用矩阵运算和快速幂算法解决特定类型的斐波那契数求和问题的方法，具体包括输入参数N、X和Y，输出为序列项平方和，并在结果过大时进行模运算处理。

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 <= 2 ³¹ – 1
X : 2<= X <= 2 ³¹– 1
Y : 2<= Y <= 2 ³¹ – 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;
}