hau1021 Fibonacci Again（递归）

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 72329    Accepted Submission(s): 33039

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

 

//(a+b)%c=(a%c+b%c)%c;
#include<iostream>
#include<math.h>
#include<stdio.h>
using namespace std;
long long a[];
int main()
{
  long long n;
  a[]=%,a[]=%;
  for(int i=;i<;i++)
    a[i]=(a[i-]+a[i-])%;//和DP很像
  while(cin>>n)
  {
    if(a[n]==)
      cout<<"yes"<<endl;
    else
      cout<<"no"<<endl;
  }
  return ;

}