hdu 1021 Fibonacci Again(同余模定理+斐波那契数列)

原创于 2015-04-15 18:11:17 发布 · 837 阅读

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#C++ #数论 #斐波那契 #同余模定理

本文探讨了一种特殊的斐波那契数列F(n)，其中F(0)=7, F(1)=11，并且对于n>2，F(n)=F(n-1)+F(n-2)。重点分析了如何判断F(n)是否能被3整除，通过计算F(n)模3的结果来简化问题。

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Fibonacci Again

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42136 Accepted Submission(s): 20151

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

题目分析：
递推的时候模3，然后判断时只有f[i]==0的时候才能被整除

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define MAX 1000007

using namespace std;

int f[MAX];
int n;

int main ( )
{
    f[0] = 7%3 , f[1] = 11%3;
    for ( int i = 2 ; i <= 1000000 ; i++ )
        f[i] = (f[i-1] + f[i-2] )%3;
    while ( ~scanf ( "%d" , &n ) )
        if ( f[n] ) puts ( "no" );
        else puts ( "yes" );
}