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
源代码：
#include<iostream> using namespace std; int main(){ int n,arr[50],i; //找出规律 // arr[0]=7;arr[1]=11; // for(i=2;i<50;i++){ // arr[i]=arr[i-1]+arr[i-2]; // arr[i]=arr[i]%3; // } // // for(i=0;i<50;i++){ // if(arr[i]%3==0)cout<<"yes"<<" "; // else cout<<"no"<<" "; // cout<<i<<" "<<arr[i]<<endl; // } while(cin>>n ){ if(n%4==2)cout<<"yes"<<endl; else cout<<"no"<<endl; } return 0; }