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

Sample Output

yes

//(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 ;

}

巴特西

hau1021 Fibonacci Again（递归）

Fibonacci Again

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