The GCD of Fibonacci Numbers

最新推荐文章于 2021-12-14 16:46:26 发布

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该博客讨论了如何计算斐波那契数列中两个特定项（fm 和 fn）的最大公约数（GCD）。它提到了一个实例，并指出解决此问题的关键在于找到第 gcd(n, m) 项的斐波那契数。" 130556579,10448902,txtai嵌入式索引详解,"['Python', '数据存储', '搜索引擎', '自然语言处理']

链接：https://ac.nowcoder.com/acm/contest/3674/A

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 ith 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).

输入描述:

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)

输出描述:

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

示例1

输入

复制

输出

复制

一开始俺以为要用矩乘写，没想到看到别人的代码原来只要求出第gcd(n,m)项斐波那契数列就行

代码：

#include<bits/stdc++.h>
using namespace std;
int T;
long long a,b,f[110];
long long gcd(long long a,long long b){
    if (!b) return a;
    return gcd(b,a%b);
}
int main(){
    f[0]=0;
    f[1]=1;
    for(int i=2;i<=45;i++) 
        f[i]=f[i-1]+f[i-2];
    scanf("%d",&T);
    while (T--){
        scanf("%lld%lld",&a,&b);
        printf("%lld\n",f[gcd(a,b)]);
    }
}