HDU1021-Fibonacci Again，，找规律就好了~~~

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 50096    Accepted Submission(s): 23727

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

题目意思不难懂吧，，a[n]%3==0？“yes\n”:"no\n";,,可是数据高达1000000，，常规斐波那契数列到40的样子就超了，，不妨打表到40看看输出结果：

for(i=0;i<45;i++)

if(a[i]%3==0) printf("%d %d  yes\n",i,a[i]);

else   printf("%d %d  no\n",i,a[i]);

看看输出结果认真观察发现了没，， 在i=2, 6, 10,14,18,22,26,30,34,38的时候输出yes,，，

没错，，if(n%4==2） printf("yes\n);

这样连数组都不用开了，，代码就不用贴出来了吧，，核心规律等都已呈上；