HDU ACM 1021 Fibonacci Again

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22501    Accepted Submission(s): 10880

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

 

Recommend

JGShining

 

#include<stdio.h>
int main()
{
    int men[8], i, j, n;
    men[0] = 1;
     men[1] = 2;
     men[2] = 0;
     men[3] = 2;
     men[4] = 2;
     men[5] = 1;
     men[6] = 0;
     men[7] = 1;
    while(scanf("%d", &n) != EOF)
    {
        if(!men[n%8]) printf("yes\n");
        else printf("no\n");
    }
} 

解题报告：
　　这题之前有尝试去AC他，不过一直都对题目的意思不理解，后来尝试解决时在几分钟内就AC了它，能得到的收获是冷静地分析数据，最重要的是草稿纸要干净，找到了规律就要大胆的验证，之前很少会这样做题，说实话这题是有点怀着吊儿郎当的心态提交的，后来想想也是有根据的，毕竟，规律就是这样的。