杭电acm1021 hdu-acm-1021解题报告

题目链接：

点击打开题目链接

题目内容：

Fibonacci Again Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 36393    Accepted Submission(s): 17580 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

题目大意：

看到题目，想到中国剩余定理，感觉这个题目一定是个有规律的；

n	0	1	2	3	4	5	6
f(n)	7	11	18	29	47	76	123
结果	no	no	yes	no	no	no	yes

..............

代码如下：

#include<stdio.h>
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		if(n%4==2)
		printf("yes\n");
		else
		printf("no\n");
	}
	return 0;
	
}