hdu1021 数学题 Fibonacci Again

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本文介绍了一种特殊的Fibonacci序列，该序列从F(0)=7和F(1)=11开始，并遵循F(n)=F(n-1)+F(n-2)的递推关系。通过一个简单的C语言程序实现，该程序预先计算了大量项并对输入值n判断F(n)是否能被3整除。

Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42109 Accepted Submission(s): 20133

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

Sample Output

no
no
yes
no
no
no

Author

Leojay

还是数学题，用到公式 (a+b)%c=(a%c+b%c)%c;

#include <stdio.h>
typedef long long ll;
const int maxn=999999+1000;
ll f[maxn];
int main()
{
    f[0]=7%3;
    f[1]=11%3;
    for(int i=2;i<maxn-100;i++)
    {
        f[i]=((f[i-1])+f[i-2])%3;
    }
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(f[n]==0)
        printf("yes\n");
    else
        printf("no\n");
    }
    return 0;
}