HDOJ 1021 Fibonacci Again

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 43372 Accepted Submission(s): 20712



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

解法一：同余定理，数列打表。

由于题中给出的数据较大，用数组储存会溢出，所以在打表时要用上同余定理。这个不多说，代码如下：

#include<stdio.h>
#define max 1000010
int a[max];
int main()
{
	int n,i;
	a[0]=7;a[1]=11;
	for(i=2;i<max;i++)
	   a[i]=((a[i-1]%3)+(a[i-2]%3))%3;//同余定理，防止数据溢出 
	while(scanf("%d",&n)!=EOF)
	{
		if(!a
)
		   printf("yes\n");
		else
		   printf("no\n");
	}
	return 0;
}

解法二：找规律。

数及对应的规律用（int型表示）
0 1 2 3 4 5 6 7 ......41 42
7 11 18 29 47 76 123 199......1568397607 负值（超出最大整数值，溢出）
数字规律
前几位数及其余数
0 1 2 3 4 5 6 7 8
数 7 11 18 29 47 76 123 199 322
余数 1 2 0 2 2 1 0 1 1
可知观察每四个为一组中第三个为yes即 i%4==2或(i-2)%4==0求即可

具体代码如下：

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