hd 1021 Fibonacci Again 规律对4取余为2 则yes

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36326    Accepted Submission(s): 17543

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

012345

 

 

Sample Output

nonoyesnonono

 

 

Author

Leojay

 

 

Recommend

JGShining   |   We have carefully selected several similar problems for you:  1061 1019 1108 1071 1049 

 

 

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