Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37229    Accepted Submission(s):
17970

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

算法分析：

一开始用的__int64 开数组挂了，估计也会挂，n的范围可以大到1000000， 如果n的值比较大，那么f[n] 就超数据类型了。

查到了这个公式：

（m+n）%3 = (m%3+n%3)%3 ;

取完余数后再存到数组里，f[n]就不会超数据类型了。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int f[1000001];

int main()
{
    int n;
	int i, j;
    f[0]=7%3 ;
	f[1]=11%3 ;
	for(i=2; i<1000000; i++)
	{
		f[i]=(f[i-1]%3+f[i-2]%3)%3;
	}
	while(scanf("%d", &n)!=EOF)
	{
		if(f[n]==0)
			printf("yes\n");
		else
			printf("no\n");
	}
	return 0;
}