HDOJ 1021 Fibonacci Again

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43818    Accepted Submission(s): 20936

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
   
   
   
   

 

Author

Leojay

 

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思路：此题若按照题目规律做，容易出现超时问题，因为终止数过大。所以要重新发现规律。
规律如下：
2，6，10，14，18，22，26。。。等每隔四个数 输出一个yes

#include<stdio.h>
int main(){
    int i,n;
    long long f[100000];
    while(scanf("%d",&n)!=EOF){
        if((n-2)%4==0) printf("yes\n");
        else   printf("no\n");
    }
    return 0;
}