剑指offer——面试题10：斐波那契数列

个人答案：

 #include"iostream"
 #include"stdio.h"
 #include"string.h"
 using namespace std;
 typedef long long ll;
 const int MAXN=;

 ll fib[MAXN];
 ll Fibonacci(int n)
 {
     if(fib[n]!=-)
         return fib[n];
     return fib[n]=Fibonacci(n-)+Fibonacci(n-);
 }

 int main()
 {
     int n;
     memset(fib,-,sizeof(fib));
     fib[]=;
     fib[]=;
     while(cin>>n)
     {
         cout<<Fibonacci(n)<<endl;
     }
     return ;
 }

官方答案：

 // 面试题10：斐波那契数列
 // 题目：写一个函数，输入n，求斐波那契（Fibonacci）数列的第n项。

 #include <cstdio>

 // ====================方法1：递归====================
 long long Fibonacci_Solution1(unsigned int n)
 {
     if(n <= )
         return ;

     if(n == )
         return ;

     return Fibonacci_Solution1(n - ) + Fibonacci_Solution1(n - );
 }

 // ====================方法2：循环====================
 long long Fibonacci_Solution2(unsigned n)
 {
     int result[] = {, };
     if(n < )
         return result[n];

     long long  fibNMinusOne = ;
     long long  fibNMinusTwo = ;
     long long  fibN = ;
     for(unsigned int i = ; i <= n; ++ i)
     {
         fibN = fibNMinusOne + fibNMinusTwo;

         fibNMinusTwo = fibNMinusOne;
         fibNMinusOne = fibN;
     }

      return fibN;
 }

 // ====================方法3：基于矩阵乘法====================
 #include <cassert>

 struct Matrix2By2
 {
     Matrix2By2
     (
         long long m00 = ,
         long long m01 = ,
         long long m10 = ,
         long long m11 =
     )
     :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 > );

     Matrix2By2 matrix;
     if(n == )
     {
         matrix = Matrix2By2(, , , );
     }
     else if(n %  == )
     {
         matrix = MatrixPower(n / );
         matrix = MatrixMultiply(matrix, matrix);
     }
     else if(n %  == )
     {
         matrix = MatrixPower((n - ) / );
         matrix = MatrixMultiply(matrix, matrix);
         matrix = MatrixMultiply(matrix, Matrix2By2(, , , ));
     }

     return matrix;
 }

 long long Fibonacci_Solution3(unsigned int n)
 {
     int result[] = {, };
     if(n < )
         return result[n];

     Matrix2By2 PowerNMinus2 = MatrixPower(n - );
     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);
     else
         printf("Test for %d in solution1 failed.\n", n);

     if(Fibonacci_Solution2(n) == expected)
         printf("Test for %d in solution2 passed.\n", n);
     else
         printf("Test for %d in solution2 failed.\n", n);

     if(Fibonacci_Solution3(n) == expected)
         printf("Test for %d in solution3 passed.\n", n);
     else
         printf("Test for %d in solution3 failed.\n", n);
 }

 int main(int argc, char* argv[])
 {
     Test(, );
     Test(, );
     Test(, );
     Test(, );
     Test(, );
     Test(, );
     Test(, );
     Test(, );
     Test(, );
     Test(, );
     Test(, );

     Test(, );

     return ;
 }