HDU1021(模运算)

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 52548    Accepted Submission(s): 24840

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

 

思路：打表函数每项模3的值，若结果为0则可以被3整除。

#include <cstdio>
using namespace std;
const int MAXN=;
int f[MAXN];
int main()
{
    f[]=%;
    f[]=%;
    for(int i=;i<;i++)
    {
        f[i]=(f[i-]+f[i-])%;
    }
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(f[n]==)
        {
            printf("yes\n");
        }
        else
        {
            printf("no\n");
        }
    }
    return ;
}