波拉契尔数列 C++

题目：写一个函数，输入n, 求斐波那契数列的第n项。

$f(n)=\left\{\begin{matrix} 0, n=0 & & & \\ 1, n=1& & & \\ f(n-1)+f(n-2),n>1& & & \end{matrix}\right.$

分析：该题有两种实现方式递归或循环。当n比较大的时候f(n)结果也会比较大，故定义的时候可以采用long(int 也行)。递归会有大量的重复计算，而循环可以把f(n-1)和f(n-2)的值存起来，避免过多的重复计算。

上代码，先递归再循环。

#include<iostream>#include<cstdlib>#include <stdio.h>#include<time.h>long Fibonacci(unsigned int n){if (n == 0)return 0;if (n == 1)return 1;return Fibonacci(n - 2) + Fibonacci(n - 1);}int main(){clock_t start, end;int n = 40;start = clock();int result=Fibonacci(n);end = clock();double time = (end - start)/ CLOCKS_PER_SEC;printf("%d %f", result,time);system("pause");}

输出结果为

#include<iostream>#include <stdio.h>#include<time.h>long Fibonacci(unsigned int n){if (n == 0)return 0;if (n == 1)return 1;long FibsubOne = 1;long FibsubTwo = 0;long Fib=0 ;int i = 2;while (i <= n){Fib = FibsubOne + FibsubTwo;FibsubTwo = FibsubOne;FibsubOne = Fib;++i;}return Fib;}int main(){clock_t start, end;int n =40;start = clock();int result=Fibonacci(n);end = clock();double time = (end - start)/ CLOCKS_PER_SEC;printf("%d %f", result,time);system("pause");}

结果显示