hdu1021 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
分析：f(n)能否被3整除.因为直接做的，数据会比较大，所以这里可以直接看余数。即f(0)=1,f(1)=2,f(3)=0,f(4)=2,f(5)=2,f(6)=1,f(7)=0,f(8)=1,f(9)=1,f(10)=2
从这儿就可以看出规律了！

#include<iostream>#include<cstring>#include<algorithm>#include<string>using namespace std;int a[]={1,2,0,2,2,1,0,1},n;int main(){    while(cin>>n)    {        int t=n%8;        if(a[t]!=0)        cout<<"no"<<endl;        else        cout<<"yes"<<endl;     }    return 0;}