我一直相信这道题有十分巧妙的解法的,去搜了好多题解发现有的太过玄妙不能领会. 最简单的就是枚举n的所有约数,然后二重循环找lcm(a, b) = n的个数 #include <cstdio> #include <vector> #include <algorithm> using namespace std; ? a : gcd(b, a % b); } int lcm(int a, int b) { return a / gcd(a, b) * b; } int ma…

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…

看题传送门 题目大意: 输入两个数G,L找出两个正整数a 和b,使得二者的最大公约数为G,最小公倍数为L,如果有多解,输出a<=b且a最小的解,无解则输出-1 思路: 方法一: 显然有G<= a <=b <=L成立.题目要求(a<=b) 所以如果有解,a最小值只能为G. a * b = G * :L所以 b = L 什么时候无解呢? 如果L 不能整除 G 就无解了嘛. #include<cstdio> int main() { int T; scanf("…

Problem F LCM Cardinality Input: Standard Input Output: Standard Output Time Limit: 2 Seconds A pair of numbers has a unique LCM but a single number can be the LCM of more than one possible pairs. For example 12 is the LCM of (1, 12), (2, 12), (3,4)…

A pair of numbers has a unique LCM but a single number can be the LCM of more than one possiblepairs. For example 12 is the LCM of (1, 12), (2, 12), (3,4) etc. For a given positive integer N, thenumber of different integer pairs with LCM is equal to N…

LCM Cardinality Input: Standard Input Output: Standard Output Time Limit: 2 Seconds A pair of numbers has a unique LCM but a single number can be the LCM of more than one possible pairs. For example 12 is the LCM of (1, 12), (2, 12), (3,4) etc. For a…

LCM Cardinality Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Description   Problem FLCM CardinalityInput: Standard Input Output: Standard Output Time Limit: 2 Seconds A pair of numbers has a unique LCM but a si…

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…

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…