ACM: 数学规律题 数论题 hdoj 1021

 

                 Fibonacci Again

Problem Description

  There are another kind ofFibonacci numbers: F(0) = 7, F(1) = 11, F(n) = F(n-1) + F(n-2)(n>=2).

 

Input

  Input consists of a sequenceof lines, each containing an integer n. (n <1,000,000).

 

Output

  Print the word "yes" if 3divide evenly into F(n).

Print the word "no" if not.

 

Sample Input

0 1 2 3 45

Sample Output

nonoyesnonono

题意: 斐波纳契数 求第n个斐波纳契数是否被3整除.

解题思路:

        1. 是否可以被整除. 可以想到: a = a (mod m);

        2. 假设:  a = b (mod m), c = d (mod m);

             ===> a*t1 - b = m*q1,   c*t2 - d = m*q2;

             ===> 两式相加: (a+c)*t - (b+d) = m*q;

             ===> 得: (a+c) = (b+d) (mod m);

        3. 可以得出结论: 只要有一个式子满足即可.

          index: 0   1   2   3   4   5   6   7   8   9   10   11   12   13   14

                 7   11  18  29  47  76  123 199 322 521 843 1364 2207  3571 5778

  a[index]%3==0: n   n   y   n   n   n   y   n   n   n   y    n    n    n    y

          取出可以整除3的下标: 2 6 10 14 ...

        4. 显然: (i-2)%4 == 0 就是可以整除3.

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;

int n;

int main()
{
// freopen("input.txt","r",stdin);
 while(scanf("%d",&n) != EOF)
 {
  if( (n-2)%4 == 0 )
   printf("yes\n");
  else
   printf("no\n");
 }
 return 0;
}