09_Fibonacci

《剑指offer》里面的这道题对fibonacci的常见计算方法进行了改进
#include<iostream>
using namespace std;long long Fibonacci_Solution1(unsigned int n)
{if (n <= 0)return 0;if (n == 1)return 1;return Fibonacci_Solution1(n - 1) + Fibonacci_Solution1(n - 2);
}// ====================方法2：循环====================
long long Fibonacci_Solution2(unsigned n)
{int result[2] = { 0, 1 };if (n < 2)return result[n];long long  fibNMinusOne = 1;long long  fibNMinusTwo = 0;long long  fibN = 0;for (unsigned int i = 2; i <= n; ++i){fibN = fibNMinusOne + fibNMinusTwo;//an=a(n-1)+a(n-2)fibNMinusTwo = fibNMinusOne;//注意这里的两句之所以看起来有些奇怪，是因为我们手算的时候都是一轮计算两个数，而这里的循环中一轮只计算1个数//这一句是an-1//fibNMinusOne = fibN;//an//以上两句是为了下一轮计算a(n+1)来作准备的}return fibN;
}// ====================方法3：基于矩阵乘法====================
#include <cassert>struct Matrix2By2
{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;
}// ====================测试代码====================
void Test(int n, int expected)
{if (Fibonacci_Solution1(n) == expected)printf("Test for %d in solution1 passed.\n", n);elseprintf("Test for %d in solution1 failed.\n", n);if (Fibonacci_Solution2(n) == expected)printf("Test for %d in solution2 passed.\n", n);elseprintf("Test for %d in solution2 failed.\n", n);if (Fibonacci_Solution3(n) == expected)printf("Test for %d in solution3 passed.\n", n);elseprintf("Test for %d in solution3 failed.\n", n);
}int main()
{Test(0, 0);Test(1, 1);Test(2, 1);Test(3, 2);Test(4, 3);Test(5, 5);Test(6, 8);Test(7, 13);Test(8, 21);Test(9, 34);Test(10, 55);Test(40, 102334155);return 0;
}
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