hdu 1021 Fibonacci Again(找規律)
阿新 • • 發佈:2018-11-25
題目:
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
解題:
程式碼:
// // main.cpp // hdu1021 // // Created by zhan_even on 2018/10/22. // Copyright © 2018年 zhan_even. All rights reserved. // #include <iostream> using namespace std; int main(int argc, const char * argv[]) { int number; while(cin >> number){ if ((number+2)%4==0 ){ cout<<"yes"<<endl; } else cout<<"no"<<endl; } return 0; }