Function解题报告

题目摘要：Define a functionf(n)=(f(n-1)+1)/f(n-2). You already got f(1) and f(2). Now, give you a numberm, please find the value of f(m).
题目大意：给出f(1)和f(2)以及m的值，根据给出的函数关系，求f(m)

输入输出要求：

Input

Thereare several test cases. Each case contains three integers indicating f(1), f(2)and m ( 1 <= f(1), f(2), m <= 1000,000,000).

 

Output

For each case, please output the value off(m), rounded to 6 decimal places.

 

输入输出样例：

Sample Input

1 1 3

 

Sample Output

2．000000

 

解题思路：可以找出规律，函数值5个一循环

代码：

#include<stdio.h>

double a,b;

int n;

double f(int m)

{

 

       if(m==1)

              returna;

       elseif (m==2)

              returnb;

       else

              return(f(m-1)+1)/f(m-2);

}

main()

{    

       doublex;

       while(scanf("%lf%lf%d",&a,&b,&n)==3)

       {

              if(n%5==0)

                     n=5;

              else

                     n=n%5;

              x=f(n);

              printf("%.4lf\n",x);

       }

       return0;

}

解题感想：这种题目很容易想到递归，最开始的思路也是用递归。但是数据太大结果就出错，原因是递归中存在除法。好在这题可以找出规律，函数值5个一循环，只要确定了f(1)和f(2)的值，f(m)就很容易可以得到。