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  • 1021 Fibonacci Again (hdoj)

    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
     
    1 #include<iostream>/*打表找规律*/
    2 using namespace std;
    3 int main(){
    4    int n;
    5    while(cin>>n){
    6         if(n%4==2) cout<<"yes"<<endl;
    7         else       cout<<"no"<<endl;
    8    }return 0;
    9 }
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  • 原文地址:https://www.cnblogs.com/a1225234/p/4496184.html
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