HDU1021

Fibonacci Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 61756 Accepted Submission(s): 28836

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

Author
Leojay

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**题目分析：
简单的数论题，f（n）要能整除3，公式可改写为F(n) = （F(n-1) + F(n-2)）mod 3，计算一下发现第二个后面的以4为基数开始循环**

#include<cstdio>int main(){    int n;    while (scanf("%d",&n)!=EOF)    {        if ((n - 2) % 4 != 0)            printf("no\n");        else            printf("yes\n");    }    return 0;}