The GCD of Fibonacci Numbers

题目

The Fibonacci sequence is the sequence of numbers such that every element is equal to the sum of the two previous elements, except for the first two elements f0 and f1 which are respectively zero and one. The first few numbers in the Recaman’s Sequence is 0,1,1,2,3,5,8,… The i-th Fibonacci number is denoted fi.

The largest integer that divides both of two integers is called the greatest common divisor of these integers. The greatest common divisor of a and b is denoted by gcd(a,b).

Two positive integers m and n are given,you must compute the GCD(fm,fn).

Input

The first linecontains one integer T（T ≤ 100）,the number of test cases.
For each test case,there are two positive integers m and n in one line. (1 ≤m,n ≤ 231 , GCD(m,n) ≤ 45)

Output

Foreach the case, your program will outputthe GCD(fm,fn).

Sample Input

Input

4
1 2
2 3
2 4
3 6

Output

1
1
1
2

思路

自己做的时候知道不可能暴力,看着求Fibonacci和Gcd差不多,找规律没找到~
                              $f\left( x\right) =f\left( x-1\right) +f\left( x-2\right)$
所以 当 $m>n$时,有$f\left( m\right) =f\left( n+1\right) f\left( m-n\right) +f\left( m-n-1\right) f\left( n\right)$  (自己推一下就可以出来撒)
所以有$Gcd\left( f\left( m\right) ,f\left( n\right) \right) =Gcd\left(f\left( n+1\right) f\left( m-n\right) +f\left( m-n-1\right) f\left( n\right),f\left( n\right) \right)$
由求$gcd$的更相减损术可知:$Gcd\left( f\left( m\right) ,f\left( n\right) \right) =Gcd\left(f\left( n+1\right) f\left( m-n\right) ,f\left( n\right) \right)$
                                                               进而可得:$=Gcd\left(f\left( n+1\right) ,f\left( n\right) \right)*Gcd\left(f\left( m-n\right) ,f\left( n\right) \right)$
                                                                              $=Gcd\left(f\left( m-n\right) ,f\left( n\right) \right)$

所以此过程中$m$和$n$互减,直到$m\%n==0$时,对应的$f(n)$就是答案,所以求$m$和$n$的最小公倍数,然后求其Fibonacci即可啦~
即$Gcd\left( f\left( m\right) ,f\left( n\right) \right) =Gcd\left(f\left( m-n\right) ,f\left( n\right) \right)=......=f\left( Gcd\left( m,n\right) \right)$
$Ps:f(n+1)和f(n)一定是互质的$

代码如下

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<climits>
#include<cctype>
#include<queue>
#include<set>
using namespace std;
typedef unsigned long long ll;
const int INF=0x3f3f3f3f;
const int mx=2e5+10;

int t;
ll n,m,s[50];

ll fibo(ll a)
{
	if(s[a])
	return s[a];
	if(a==0)
	return 0;
	if(a==1)
	return 1;
	return s[a]=fibo(a-1)+fibo(a-2);
}

ll gcd(ll n,ll m)
{
	return n%m==0?m:gcd(m,n%m);
}

int main()
{
	fibo(50);
	s[0]=0,s[1]=1;
	int t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%lld%lld",&n,&m);
		printf("%lld\n",s[gcd(n,m)]);
	}
	return 0;
}

来源：CSDN

作者：楚仙子

链接：https://blog.csdn.net/weixin_43901733/article/details/103949097