HDU 1021 Fibonacci Again

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
    
    
    
    
 
    
    
    
    
题意就是判断a[n]是否为3 的倍数。是，输出yes，否则，输出no。
    
    
    
    
若直接相加数据就会很大，每个数都直接对3取余再运算，这样a[n]的值只在0 1 2之间徘徊
    
    
    
    
 
    
    
    
    
#include<stdio.h> #include<string.h> #include<math.h> #include<iostream> #include<algorithm> #include<string> using namespace std; __int64 a[1000001]= {1,2}; int main() {     for(int i=2; i<=1000000; i++)     {         a[i]=a[i-1]+a[i-2];         a[i]=a[i]%3;     }     int n;     while(cin>>n)     {         if(a[n]==0)             cout<<"yes"<<endl;         else             cout<<"no"<<endl;     } }