斐波那契模

斐波那契数列定义为$F_i$ = $F_{i-1}$ + $F_{i-2}$在哪里$F_1$ = 1并且$F_2$ = 1是种子。
对于给定的质数p，请考虑一个新序列，即(斐波那契序列)mod p。例如，对于p = 5，新序列将为1、1、2、3、0、3、3、1、4、0、4、4…
新序列的最小零定义为第一个斐波那契数，它是p或p的倍数。 $F_i$ mod p = 0。
给定素数p，找到序列Fibonacci模p的最小零。
例子：

Input : 5
Output : 5
The fifth Fibonacci no (1 1 2 3 5) 
is divisible by 5 so 5 % 5 = 0.

Input : 7
Output : 8
The 8th Fibonacci no (1 1 2 3 5 8 13 21) 
is divisible by 7 so 21 % 7 = 0.

一种简单的方法是继续计算斐波那契数，并为每个数计算Fi mod p。但是，如果我们观察到这个新序列，则让$r_i$表示$i_th$序列的项，则如下： $r_i$ =( $r_{i-1}$ + $r_{i-2}$ )mod pie其余部分$r_i$实际上是本系列前两个术语余数的总和。因此，我们无需生成斐波那契数列，然后对每个项取模，我们只需将前两个余数相加，然后取其模数p。
下面是查找最小值0的实现。

C++

// C++ program to find minimal 0 Fibonacci
// for a prime number p
#include
using namespace std;
 
// Returns position of first Fibonacci number
// whose modulo p is 0.
int findMinZero(int p)
{
    int first = 1, second = 1, number = 2, next = 1;
    while (next)
    {
        next = (first + second) % p;
        first = second;
        second = next;
        number++;
    }
    return number;
}
 
// Driver code
int main()
{
    int p = 7;
    cout << "Minimal zero is: "
        << findMinZero(p) << endl;
    return 0;
}

Java

// Java program to find minimal 0 Fibonacci
// for a prime number p
import java.io.*;
 
class FibZero
{
    // Function that returns position of first Fibonacci number
    // whose modulo p is 0
    static int findMinZero(int p)
    {
        int first = 1, second = 1, number = 2, next = 1;
        while (next > 0)
        {
            // add previous two remainders and
            // then take its modulo p.
            next = (first + second) % p;
            first = second;
            second = next;
            number++;
        }
        return number;
    }
     
    // Driver program
    public static void main (String[] args)
    {
        int p = 7;
        System.out.println("Minimal zero is " + findMinZero(p));
    }
}

Python3

# Python 3 program to find minimal
# 0 Fibonacci for a prime number p
 
# Returns position of first Fibonacci
# number whose modulo p is 0.
def findMinZero(p):
    first = 1
    second = 1
    number = 2
    next = 1
 
    while (next):
        next = (first + second) % p
        first = second
        second = next
        number = number + 1
     
    return number
 
# Driver code
if __name__ == '__main__':
    p = 7
    print("Minimal zero is:", findMinZero(p))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find minimal 0
// Fibonacci for a prime number p
using System;
 
class GFG {
     
    // Function that returns position
    // of first Fibonacci number
    // whose modulo p is 0
    static int findMinZero(int p)
    {
        int first = 1, second = 1;
        int number = 2, next = 1;
        while (next > 0)
        {
             
            // add previous two
            // remainders and then
            // take its modulo p.
            next = (first + second) % p;
            first = second;
            second = next;
            number++;
        }
        return number;
    }
     
    // Driver program
    public static void Main ()
    {
        int p = 7;
        Console.WriteLine("Minimal zero "
              + "is :" + findMinZero(p));
    }
}
 
// This code is contributed by anuj_67.

Javascript

// Javascript program to find
// minimal 0 Fibonacci
// for a prime number p
 
// Returns position of
// first Fibonacci number
// whose modulo p is 0.
function findMinZero(p)
{
    let first = 1;
    let second = 1;
    let number = 2;
    let next = 1;
    while (next)
    {
        next = (first +
                second) % p;
        first = second;
        second = next;
        number++;
    }
 
    return number;
}
 
// Driver code
let p = 7;
document.write("Minimal zero is: ",
    findMinZero(p) + "<br>");
 
// This code is contributed
// by akt_mit

输出：

Minimal zero is: 8