HDOJ-1021

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 50987    Accepted Submission(s): 24142


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
 

本题数据较大,故不适合用递推来解决,既然不能递推,我们很自然的就会想到找规律。

根据题目已知条件:
Print the word”yes” if 3 divide evenly into F(n);Print the word”no” if not.
这里mod取值为3,则可将公式条件演变为:
综上所述,可得到以下对应关系:F(0)= 1, F(1) = 2, F(n) = ( F(n-1) + F(n-2)  )( mod 3) (n>=2).
index  0  1  2  3  4  5  6  7  8  9  10  11  12  13
value  1  2  0  2  2  1  0  1  1  2   0   2   2  1
print  no no yes no  no no yes  no  no  no  yes  no  no  no
这样我们就得到了如下规律:从第2个开始每隔4个循环一次。

附AC代码:

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cstring>
 4 using namespace std;
 5 int main(){
 6     int n;
 7     while(~scanf("%d",&n)){
 8         if((n-2)%4)
 9         printf("no\n");
10         else
11         printf("yes\n");
12     }
13     return 0;
14 }

 

posted @ 2016-05-11 23:09  Kiven#5197  阅读(222)  评论(0编辑  收藏  举报