hdoj1021Fibonacci Again

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第二个方法是找出的规律。你们可以观察观察。

/*Fibonacci Again
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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*/

<span style="font-size:18px;"># include<stdio.h># define  m 1000001int f[m];int main(){        int n,i;        f[0]=7;        f[1]=11;        for(i=2;i<=m;i++)        f[i]=(f[i-1]%3+f[i-2]%3)%3;        while(scanf("%d",&n)!=EOF)    {        if(f[n]%3==0)        printf("yes\n");        else        printf("no\n");    }    return 0;}</span>

<span style="font-size:18px;">#include<stdio.h> main() {  int n;  while(~scanf("%d",&n))  {   if(n%4==2) printf("yes\n");   else printf("no\n");  } }</span>

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