Fibonacci Again 分类： HDU 2015-06-26 11:05 13人阅读 评论(0) 收藏

Fibonacci Again
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43539 Accepted Submission(s): 20797

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

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>

using namespace std;

const int Max=1100000;
int Arr[Max];
int main()
{
    memset(Arr,0,sizeof(Arr));
    Arr[0]=1;
    Arr[1]=2;
    for(int i=2;i<Max;i++)
    {
        Arr[i]=(Arr[i-1]+Arr[i-2])%3;
    }
    int n;
    while(~scanf("%d",&n))
    {
        if(!Arr[n])
        {
            printf("yes
");
        }
        else
        {
            printf("no
");
        }

    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。