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

Sample Output

yes

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：

 1 #include <cstdio>
 2 #include <iostream>
 3 using namespace std;
 4 int main()
 5 {
 6     int n;
 7     while(~scanf("%d",&n))
 8     {
 9         if((n-2)%4==0)
10             printf("yes
");
11         else
12             printf("no
");
13     }
14     return 0;
15 }