hdu 1021 Fibonacci Again

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44959    Accepted Submission(s): 21451

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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简单的找规律，打表后发现每4个一次循环。

 

题意：按题意 F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2) 这个公式求值，判断这个F[n]是否是3的倍数，是输出yes，不是输出no。

 

附上代码：

 

#include <cstdio>
#include <iostream>
using namespace std;
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        if((n-2)%4==0)
            printf("yes\n");
        else
            printf("no\n");
    }
    return 0;
}