hdu 1021 Fibonacci Again（斐波那契数列，取余）

最新推荐文章于 2021-02-27 04:48:44 发布

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最新推荐文章于 2021-02-27 04:48:44 发布

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分类专栏：数学-数论文章标签： ACM 数论斐波那契数列

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本文探讨了ACM问题HDU 1021中关于斐波那契数列与取余操作的关系，通过代码展示如何利用数论技巧优化计算过程。

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Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 51621 Accepted Submission(s): 24423

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

思路：边计算边取余即可，取余之后贡献值0,1,2不会改变

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define LL long long
#define N  1000010
LL f[N];
int main()
{
    int n;
    LL a,b;
    f[0]=7%3,f[1]=11%3;
    while(~scanf("%d",&n))
    {
        for(int i=2; i<=n; i++)
            f[i]=(f[i-2]+f[i-1])%3;
            if(f[n]) printf("no\n");
            else printf("yes\n");
    }
    return 0;
}