HDU1021Fibonacci Again

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 49607    Accepted Submission(s): 23534

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：

它新定义了一个斐波那契数列为 F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2) (n>=2)；

输入有多行，每一行有一个正整数  n（n<1000000）；

输出：

Print the word "yes" if 3 divide evenly into F(n).

就是说F[n]能够被3整除，也就是F[n]对3去余为0；

思路：

如果有正整数 a,  b,  c,   d, x; 假如: c=a+b,     x=c%d ,那么可以知道  x=(a+b)%d;  易推知，x=(a%d+b%d)%d;

ps:参考同余定理

给出AC代码：

#include<iostream>
using namespace std;
int F[1000005];
int main()
{
	F[0] = 7 % 3, F[1] = 11 % 3;
	for (int i = 2; i < 1000005; i++)
		F[i] = (F[i - 1] + F[i - 2]) % 3;
	int n;
	while (cin >> n)
	{
		if (F[n] == 0)cout << "yes" << endl;
		else cout << "no" << endl;
	}
	return 0;
}