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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

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

This function is a good one I accidentally saw. It is awesome and I like it.
Is used to find the minimum public approx.
A simple description is that gcd (a, B) indicates the maximum public factor of non-negative integers A and B, so: gcd (a, B) =

Question link:
HuangJingQuestion:
Find the GCD with the K value of two numbers
Ideas:
First, find the maximum common divisor. My first thought was to create a large prime number table, and then continue division to find the largest K number, but

First, we will introduce two:
GCD with large numbers
Stein + Euclidean
function stein(a,b:int64):int64;begin if a>1,b>>1)>1,b)); if (b and 1)=0 then exit(stein(a,b>>1)); exit(stein((a+b)>>1,(a-b)>>1));end;View code
Decimal GCD
Phase

The type of the two topics feel the same is the use of the idea of the repulsion to push forward and then go heavy.HDU 5212 Test Instructions:The number of n is calculated for each of the n numbers, and F (i), f (i) = GCD (A[i], a[j]) * (GCD (a[i],

2016 Dalian Network Competition --- Different GCD Subarray Query (GCD discretization + tree array), tree array discretization
Question Link
Http://acm.split.hdu.edu.cn/showproblem.php? Pid = 1, 5869
Problem DescriptionThis is a simple problem. the

Poj 2429 GCD & LCM Inverse [java] + [mathematics], poj2429gcd
GCD & LCM Inverse
Time Limit:2000 MS
Memory Limit:65536 K
Total Submissions:9928
Accepted:1843
DescriptionGiven two positive integers a and B, we

Description
Given the value of N, you will have to find the value of G. The meaning of G is given in the following code
G=0;
for(i=1;i<N;i++)
for(j=i+1;j
G+=gcd(i,j);
/* Here gcd () is a function that finds the greatest

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