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("…
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…
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…
Least Common Multiple Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 53016 Accepted Submission(s): 20171 Problem Description The least common multiple (LCM) of a set of positive integers is…
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…
