Fibonacci Again
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 39 Accepted Submission(s) : 31
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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
Sample Output
Author
Leojay
找规律的题目。
把f（n）分裂开就能够得到规律。
AC代码：
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
long n;
while(scanf("%ld",&n) != EOF)
if (n%8==2 || n%8==6)
printf("yes\n");
else
printf("no\n");
return 0;
}
运行结果：
HDU Fibonacci Again
原文：http://blog.csdn.net/zp___waj/article/details/46041703
