求斐波那契數(shù)列的第項(xiàng)。
答案：
// ====================方法1：遞歸====================
long long Fibonacci_Solution1(unsigned int n)
{
if (n==0) return 0;
else if (n==1) return 1;
else return Fibonacci_Solution1(n-1)+Fibonacci_Solution1(n-2);
}
// ====================方法2：循環(huán)遞推====================
long long Fibonacci_Solution2(unsigned int n)
{
if (n==0) return 0;
else if (n==1) return 1;
else
{
long long prepre = 0, pre = 1;
long long ret = 0;
for (int i=2; i<=n; i++)
{
ret = pre + prepre;
prepre = pre;
pre = ret;
}
return ret;
}
}
// =================方法3：基于矩陣乘法=================
#include <cassert>
struct Matrix2By2
{
explicit Matrix2By2(long long m00 = 0, long long m01 = 0,
long long m10 = 0, long long m11 = 0)
:m_00(m00), m_01(m01), m_10(m10), m_11(m11) {}
long long m_00;
long long m_01;
long long m_10;
long long 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(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);
}
else if(n % 2 == 1)
{
matrix = MatrixPower((n - 1) / 2);
matrix = MatrixMultiply(matrix, matrix);
matrix = MatrixMultiply(matrix, Matrix2By2(1, 1, 1, 0));
}
return matrix;
}
long long Fibonacci_Solution3(unsigned int n)
{
int result[2] = {0, 1};
if(n < 2)
return result[n];
Matrix2By2 PowerNMinus2 = MatrixPower(n - 1);
return PowerNMinus2.m_00;
}
