Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 39394    Accepted Submission(s): 18895

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

012345

 

Sample Output

nonoyesnonono
#include<stdio.h> #include<stdlib.h> #include<string.h> #define N 1000000 int a[N]; int main() {  int n; memset(a,0,sizeof(a)); a[0]=7%3;a[1]=11%3; while(scanf("%d",&n)!=EOF) { for(int i=2;i<=n;i++) { a[i]=(a[i-1]%3+a[i-2]%3)%3; } if(a[n]==0) { printf("yes\n"); } else { printf("no\n"); } } }

(a + b) % p = (a % p + b % p) % p （1）求模算法