Hangzhou Electric 1021--fibonacci Again

Fibonacci Again

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 43441 Accepted Submission (s): 20755

Problem Descriptionthere is another kind of Fibonacci numbers:f (0) = 7, f (1) = one, f (n) = f (n-1) + f (n-2) (n>=2).

Inputinput consists of a sequence of lines, each containing an integer n. (n < 1,000,000).

Outputprint 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

authorleojay//7 11 18 29 47 76 123 .... Die 3 after the remainder is 1 2 0 2 2 1 0 ... , write a few groups will find as long as meet n%4==2; F[n] will be divisible by 3;

1#include <stdio.h>2 intMain ()3 {4     intN;5      while(~SCANF ("%d",&N))6     {7         if(n%4==2)8printf"yes\n");9         ElseTenprintf"no\n"); One     } A     return 0; -}

Hangzhou Electric 1021--fibonacci Again