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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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 #include2 #include 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\n");11 else12 printf("no\n");13 }14 return 0;15 }