HDU1021（规律题）

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37739    Accepted Submission(s): 18208

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

012345

 

Sample Output

nonoyesnonono

 

Author

Leojay

 

方法：

数及对应的规律用（int型表示）0 1  2  3  4  
[csharp] view plain copy
 print?
  
5  6   7  ......41         427 11 18 29 47 76 123 199......1568397607 负值（超出最大整数值，溢出）考虑溢出问题。1、直接法 f(0) = 7%3;f(1) = 11%3;i>=2时f(i) = (f(i-1)%3+f(i-2)%3)%3;2.数字规律前几位数及其余数   0 1  2  3  4  5  6   7   8 数7 11 18 29 47 76 123 199 322余数1 2  0  2  2  1  0   1   1可知观察每四个为一组中第三个为yes即 i%4==2或(i-2)%4==0求即可

[csharp] view plain copy

print?

1. #include<stdio.h>  
2. int main()  
3. {  
4.     int i,n;  
5.     while(scanf("%d",&n)!=EOF)  
6.     {  
7.         if((n-2)%4==0||n%4==2)  
8.             printf("yes\n");  
9.         else  
10.             printf("no\n");  
11.     }  
12.     return 0;  
13. }  

#include<stdio.h>int main(){int i,n;while(scanf("%d",&n)!=EOF){if((n-2)%4==0||n%4==2)printf("yes\n");elseprintf("no\n");}return 0;}