gcd(a, b),就是求a和b的最大公约数 lcm(a, b),就是求a和b的最小公倍数 然后有个公式 a*b = gcd * lcm     ( gcd就是gcd(a, b), ( •̀∀•́ ) 简写你懂吗) 解释(不想看就跳过){ 首先,求一个gcd,然后... a / gcd 和 b / gcd 这两个数互质了,也就是 gcd(   a / gcd ,b / gcd  )  =  1,然后... lcm = gcd *  (a / gcd) * (b / gcd) lcm = (a *…

数论入门2 另一种类型的数论... GCD,LCM 定义\(gcd(a,b)\)为a和b的最大公约数,\(lcm(a,b)\)为a和b的最小公倍数,则有: 将a和b分解质因数为\(a=p1^{a1}p2^{a2}p3^{a3}...pn^{an},b=p1^{b1}p2^{b2}p3^{b3}...pn^{bn}\),那么\(gcd(a,b)=\prod_{i=1}^{n}pi^{min(ai,bi)},lcm(a,b)=\prod_{i=1}^{n}pi^{max(ai,bi)}\)(0和任何…

GCD?LCM! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 316    Accepted Submission(s): 200 Output T lines, find S(n) mod 258280327. Sample Input 8 1 2 3 4 10 100 233 11037 Sample Output 1 5 1…

根据最大公约数和最小公倍数求原来的两个数 题目大意,不翻译了,就是上面链接的意思. 具体思路就是要根据数论来,设a和b的GCD(最大公约数)和LCM(最小公倍数),则a/GCD*b/GCD=LCM/GCD,我们只用枚举LCM/GCD的所有质因数就可以了,然后把相应的质因数乘以GCD即可得出答案. 找素数很简单,用Miller_Rabin求素数的方法,可以多求几次提高正确率,原理就是用的费马定理:如果P是素数,则A^(p-1)mod P恒等于1,为了绕过Carmichael数,采用费马小定理:如果…

Least Common Multiple (HDU - 1019) [简单数论][LCM][欧几里得辗转相除法] 标签: 入门讲座题解 数论 题目描述 The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7…

UVA11388 GCD LCM Description of the title PDF The GCD of two positive integers is the largest integer that divides both the integers without any remainder. The LCM of two positive integers is the smallest positive integer that is divisible by both th…

题意:给出a和b的gcd和lcm,让你求a和b.按升序输出a和b.若有多组满足条件的a和b,那么输出a+b最小的.思路:lcm=a*b/gcd   lcm/gcd=a/gcd*b/gcd 可知a/gcd与b/gcd互质,由此我们可以先用Pollard_rho法对lcm/gcd进行整数分解, 然后对其因子进行深搜找出符合条件的两个互质的因数,然后再都乘以gcd即为输出答案. #include <iostream> #include <stdio.h> #include <alg…

GCD & LCM Inverse Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10621   Accepted: 1939 Description Given two positive integers a and b, we can easily calculate the greatest common divisor (GCD) and the least common multiple (LCM) of a…

[题目链接] http://poj.org/problem?id=2429 [题目大意] 给出最大公约数和最小公倍数,满足要求的x和y,且x+y最小 [题解] 我们发现,(x/gcd)*(y/gcd)=lcm/gcd,并且x/gcd和y/gcd互质 那么我们先利用把所有的质数求出来Pollard_Rho,将相同的质数合并 现在的问题转变成把合并后的质数分为两堆,使得x+y最小 我们考虑不等式a+b>=2sqrt(ab),在a趋向于sqrt(ab)的时候a+b越小 所以我们通过搜索求出最逼近sqr…

II U C   ONLINE   C ON TEST  Problem D: GCD LCM Input: standard input Output: standard output The GCD of two positive integers is the largest integer that divides both the integers without any remainder. The LCM of two positive integers is the smalle…