九度OJ 1092:Fibonacci (递归)
- 题目描述:
- 
The Fibonacci Numbers{0,1,1,2,3,5,8,13,21,34,55...} are defined by the recurrence: 
 F0=0 F1=1 Fn=Fn-1+Fn-2,n>=2
 Write a program to calculate the Fibonacci Numbers.
- 输入:
- 
Each case contains a number n and you are expected to calculate Fn.(0<=n<=30) 。 
- 输出:
- 
For each case, print a number Fn on a separate line,which means the nth Fibonacci Number. 
- 样例输入:
- 
1 
- 样例输出:
- 
1 
思路:
用数组递归
代码:
#include <stdio.h>
 
int fib(int n)
{
    int a;
    if (n==0)
        a = 0;
    else if (n==1)
        a = 1;
    else
        a = fib(n-2) + fib(n-1);
    return a;
}
 
int main(void)
{
    int n;
 
    while (scanf("%d", &n) != EOF)
    {
        printf("%d\n", fib(n));
    }
 
    return 0;
}
/**************************************************************
    Problem: 1092
    User: liangrx06
    Language: C
    Result: Accepted
    Time:30 ms
    Memory:912 kb
****************************************************************/
    编程算法爱好者。
 
                    
                 
                
            
         
 浙公网安备 33010602011771号
浙公网安备 33010602011771号