Fibonacci Again (簡單規律題)
阿新 • • 發佈:2018-12-09
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
#include<iostream> using namespace std; int main() { int n,m,j,k,i,T; while (cin>>i) { if (i==0) cout<<"no"<<endl; else if (i==1) cout<<"no"<<endl; else { if (i%4==2) cout<<"yes"<<endl; else cout<<"no"<<endl; } } return 0; }