Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 66450 Accepted Submission(s): 30760

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

【题意】：若F(n)%3==0输出yes

【分析】：根据打表找规律发现若n%4==2直接输出yes

【打表代码】：

#include<bits/stdc++.h>

using namespace std;

int main()

{

    int a[];

    a[]=;

    a[]=;

    for(int i=;i<=;i++)

        a[i]=a[i-]%+a[i-]%;

    for(int i=;i<=;i++)

        printf("%d\n",a[i]%);

}

打出部分表

【代码】：

#include<bits/stdc++.h>

using namespace std;

int main()

{

    int n;

    while(cin>>n)

    {

        puts(n%==?"yes":"no");

    }

}

注意yes大小写！

【总结】：一般看到【类斐波那契数列】【%某个数】就想到找规律、打表

巴特西

HDU 1021 Fibonacci Again【打表找规律】

Fibonacci Again

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