HDOJ 1021 Fibonacci Again

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Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47873 Accepted Submission(s): 22734

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

no
no
yes
no
no
no

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#define N 1000000
int wqs[N+10];
int fib() {
	int i;
	wqs[0]=7%3,wqs[1]=11%3;
	for(i=2; i<=N; i++) {
		wqs[i]=(wqs[i-1]%3+wqs[i-2]%3)%3;
	}
}
int main() {
	int n;
	fib();
	while(scanf("%d",&n)!=EOF) {
		if(!wqs[n])
			printf("yes\n");
		else
			printf("no\n");
	}
	return 0;
}
/*
Fibonacci Again
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

数据量较大，预处理之后求解。利用同余定理
*/