gcd recursive

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Euclidean algorithm GCD and its development ultimate explanation

Extended Euclidean algorithm-solving indefinite equation, linear congruence equation.If the two frogs meet after the S-step, the following equations will be fulfilled:(x+m*s)-(y+n*s) =k*l (k=0,1,2 ...)Slightly changed to form:(n-m) *s+k*l=x-yMake

Euclidean Algorithm gcd and its ultimate explanation

Euclidean Algorithm gcd and its ultimate explanation This problem has plagued me for a long time. I finally found an explanation, and I made some changes myself. I will certainly be able to deepen my understanding after my patience.  Extended

Summary of GCD and ex_gcd

Gcd () --- indicates the maximum common number. The common method is Euclidean algorithm. Ex_gcd () --- Extended Euclidean Algorithm Definition 1: A and B are two integers not all 0, that is, the maximum common divisor of A and B is the maximum

Calculate the maximum common divisor gcd (A, B) of two numbers & proves Euclidean Algorithm

Calculate the maximum common divisor of two numbers A and B. We can think of enumerating each positive integer from [1, min (A, B: #includeusing namespace std;int gcd(int a,int b){ int ans=1; for(int i=2;i>a>>b; cout   However, when a and B

The oldest algorithm: Euclidean method (two natural number GCD)

In mathematics, the Euclidean algorithm, also known as Euclid, is considered to be the world's oldest algorithm (300 BC), which is used to find two GCD algorithms. The Euclidean method first appeared in Euclid's "Geometrical Original" (Volume VII,

The realization code _c language of GCD and LCM by recursive method

Mathematical principle:There are two numbers of NUM1 and num2, assuming that the NUM1 is relatively large. Make remainder r = num1% num2.When r = = 0 o'clock, that is, NUM1 can be divisible by num2, obviously num2 is the gcd of these two

Greatest common divisor (GCD) algorithm (Euclid)

One article reproduced from the three fist of the farmerEuclidean algorithm and extended Euclidean algorithmEuclidean algorithm, also known as the greatest common divisor method, is used to calculate two integers, a, b, and so on. Its computational

uva11426 GCD Extreme (II)

Test instructions: Sum (gcd (i,j), 11Ideas:1. Establish a recursive relationship, S (n) =s (n-1) +gcd (1,n) +gcd (2,n) +......+GCD (n-1,n);2. Set f (n) =gcd (1,n) +gcd (2,n) +......+GCD (n-1,n).GCD (X,n) =i is an approximate (xThe GCD (x,n) =i is

UVa 1393 (tolerant principle, GCD) highways

Test instructionsA lattice of n rows m columns is given to find out how many non-horizontal non-vertical lines pass at least two points.Analysis:First of all, the purple book on the idea, programming is simple and easy to understand. Because of

HDU acmsteps 2.1.1 least public multiple (GCD)

Minimum Public multiple Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others) Total submission (s): 2010 accepted submission (s): 1543 Problem description Given two positive integers, calculate the minimum public

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