Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 50863 Accepted Submission(s): 24079

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

一来想把数列所有值计算出来，发现数值增长实在太快，到几百的规模时longlong就溢出了。

然后自己推了一下发现，（a+b）%c==(a%c+b%c)%c，这应该是取模运算的性质。然后就过了。

#include<iostream>

#include<cstdio>

using namespace std;

#define LL long long

int f[];

int main()

{

    f[]=;

    f[]=;

    int t=;

    while(t<)

    {

        f[t]=(f[t-]+f[t-])%;

        t++;

    }

    int n;

    while(scanf("%d",&n)!=EOF)

    {

        if(f[n]==)

            printf("yes\n");

        else

            printf("no\n");

    }

    return ;

}

巴特西

HDU_1021_费布拉切变形

Fibonacci Again

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