2070 Fibbonacci Number

http://acm.hdu.edu.cn/showproblem.php?pid=2070

Fibbonacci Number

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13386    Accepted Submission(s): 6675


Problem Description
Your objective for this question is to develop a program which will generate a fibbonacci number. The fibbonacci function is defined as such:

f(0) = 0
f(1) = 1
f(n) = f(n-1) + f(n-2)

Your program should be able to handle values of n in the range 0 to 50.
 


 

Input
Each test case consists of one integer n in a single line where 0≤n≤50. The input is terminated by -1.
 


 

Output
Print out the answer in a single line for each test case.
 


 

Sample Input
  
  
3 4 5 -1
 


 

Sample Output
  
  
2 3 5
Hint
Note:
you can use 64bit integer: __int64

 

 

用递归和普通方法会超时,需要先打表

#include<stdio.h>
__int64 a[51];
int main()
{   
    int i;
    for(i=2,a[0]=0,a[1]=1;i<=51;i++)
    a[i]=a[i-1]+a[i-2];
    
    int n;
    while(scanf("%d",&n)&&n!=-1)
    {
           printf("%I64d\n",a[n]);
    }
    return 0;
}                     


 

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值