hdu-1021 Fibonacci Again

题目链接;

http://acm.hdu.edu.cn/showproblem.php?pid=1021

题目类型：

斐波那契数列

题意描述：

一个斐波那契数列，如果对3取余为0，输出yes，反之输出no。

解题思路：

先在全局变量定一个较大值Max为1000010，然后打表，然后在写多实例输入，判断输出即可。

题目：

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 60963    Accepted Submission(s): 28487

Problem Description

There are another kind of Fibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2).

 

Input

Input consists of a sequence of lines, each containing an integer n. (n < 1,000,000).

 

Output

Print the word "yes" if 3 divide evenly into F(n).

Print the word "no" if not.

 

Sample Input

0

1

2

3

4

5

 

Sample Output

no

no

yes

no

no

no

 

# include <stdio.h>
# define MAX 1000010

int a[MAX];

int main ()
{
    int i,n;
    a[0]=7%3; a[1]=11%3;
    for(i=2;i<MAX;i++)
        a[i]=(a[i-1]+a[i-2])%3;
    while(scanf("%d",&n)!=EOF)
    {
        if(a[n]==0)
            printf("yes
");
        else
            printf("no
");
    }
    return 0;
}