已知Fibonacci的矩阵乘法,即:
求Fn得最后4位。
给出的公式可以用数学归纳法证明,我没证,以后有兴趣的时候再证吧。
矩阵乘法求Fn算是最快的一种算法。参考
blog.csdn.net/yysdsyl/archive/2007/10/24/1841865.aspx
n非常大时求不来,只是要求最后几位,所以每乘一次就模10000一次。
n-1次之后,矩阵的第一行第一列就是Fn的最末几位。
求矩阵乘方时要用二分法,若n为偶,则[A]^n=[A]^(n/2)*[A]^(n/2).
若为奇,则[A]^n=[A]^((n-1)/2)*[A]^((n-1)/2)
写了七八十行。不过还是有不懂得地方:
结构体里面写了个貌似函数的东西不知什么意思,尤其是那个“:”
#include "iostream"
#include
using namespace System;
using namespace std;
struct Matrix2By2
{
Matrix2By2(int m00 = 0, int m01 = 0, int m10 = 0, int m11 = 0)
:m_00(m00), m_01(m01), m_10(m10), m_11(m11)
{
}
int m_00;
int m_01;
int m_10;
int m_11;
};
Matrix2By2 MatrixMultiply
(
const Matrix2By2& matrix1,
const Matrix2By2& matrix2
)
{
return Matrix2By2(
matrix1.m_00 * matrix2.m_00 + matrix1.m_01 * matrix2.m_10,
matrix1.m_00 * matrix2.m_01 + matrix1.m_01 * matrix2.m_11,
matrix1.m_10 * matrix2.m_00 + matrix1.m_11 * matrix2.m_10,
matrix1.m_10 * matrix2.m_01 + matrix1.m_11 * matrix2.m_11);
}
Matrix2By2 MatrixPower(long unsigned int n)
{
assert(n > 0);
Matrix2By2 matrix;
if(n == 1)
{
matrix = Matrix2By2(1, 1, 1, 0);
}
else if(n % 2 == 0)
{
matrix = MatrixPower(n / 2);
matrix = MatrixMultiply(matrix, matrix);
matrix.m_00%=10000;
matrix.m_01%=10000;
matrix.m_10%=10000;
matrix.m_11%=10000;
}
else if(n % 2 == 1)
{
matrix = MatrixPower((n - 1) / 2);
matrix = MatrixMultiply(matrix, matrix);
matrix = MatrixMultiply(matrix, Matrix2By2(1, 1, 1, 0));
matrix.m_00%=10000;
matrix.m_01%=10000;
matrix.m_10%=10000;
matrix.m_11%=10000;
}
return matrix;
}
int Fibonacci_Solution3(long unsigned int n)
{
int result[2] = {0, 1};
if(n < 2)return result[n];
Matrix2By2 PowerNMinus2 = MatrixPower(n - 1);
return PowerNMinus2.m_00;
}
int main(array ^args)
{
long signed int n,i=0;
int res;
cin>>n;
while(n!=-1){
res=Fibonacci_Solution3(n);
cout<
cin>>n;
}
return 0;
}
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