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  • (HDOJ 1021)Fibonacci Again

    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
    5
     

    Sample Output
    no 
    no 
    yes 
    no 
    no 
    no
     

    Author
    Leojay
     

    Recommend
    JGShining
     
     AC code:

    #include <iostream>
    using namespace std;
    int main()
    {
     
    int n;
     
    while(cin>>n)
     {
      
    int k = n % 8;
      
    if(k==2 || k==6)
      {
       cout
    <<"yes"<<endl;
      }
      
    else
       cout
    <<"no"<<endl;
     }
     
    return 0;
    } 

    作者:cpoint
    本文版权归作者和博客园共有,欢迎转载,但未经作者同意必须保留此段声明,且在文章页面明显位置给出原文连接,否则保留追究法律责任的权利.
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  • 原文地址:https://www.cnblogs.com/cpoint/p/2015279.html
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