HDOJ1021 Fibonacci Again

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 72792 Accepted Submission(s): 33298

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

Author

Leojay

Code

这是一道规律题,将Fibonacci数列对3取模输出后,可以发现是一个周期为8的数列.

#include <cstdio>
int main()
{
    freopen("in.txt", "r", stdin);
    int n;
    while(~scanf("%d", &n))
        if(n % 8 == 2 || n % 8 == 6) printf("yes\n");
        else printf("no\n");
    return 0;
}
/*
 * 错误的代码
#include <iostream>
using namespace std;
const int maxn = 1000000 + 5;
long long fi[maxn];
int main()
{
    freopen("in.txt", "r", stdin);
    int fin = 2;
    int n;
    fi[0] = 7;
    fi[1] = 11;
    while(~scanf("%d", &n))
    {
        if(fin <= n)
        {
            for(int i = fin; i <= n; i++)
                fi[i] = fi[i - 1] + fi[i - 2];
            fin = n;
        }
        if(fi[n] % 3 == 0)
            printf("yes\n");
        else
            printf("no\n");
    }
    return 0;
}
*/