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.
Print the word "no" if not.
Sample Input
0
1
2
3
4
5
Sample Output
no
no
yes
no
no
no
Author
Leojay
Recommend
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 }