题目链接: https://cn.vjudge.net/problem/23709/origin 本题其实有坑 数据大小太大, 2的32次方,故而一定是取巧的算法,暴力不可能过的 思路是最大公因数的倍数是最小公倍数,又有a <= b所以可以知道 a = gcd, b = lcm AC代码如下: #include <cstdio> #define ll long long using namespace std; int main() { int T; scanf("%d&quo…

两个数的最小公倍数和最大公约数肯定是倍数关系 然后又让求使得a最小  因为 a = m * gcd 令m = 1 时 a取得最小  即gcd 则b = lcm #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <queue> #include <vector> #include <cmath> #de…

代码很短理解不容易,在这说不清,大家代码里寻真相. 为什么二者相除就可以A了多找点数试试理解理解. #include<stdio.h> #define mod 1000000007 #define ll long long int main() { ll G, L,n; int t; scanf("%d",&t); while(t--) { scanf("%lld %lld",&G,&L); ) printf("%lld…

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…

组合数学 GCD and LCM Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 451    Accepted Submission(s): 216 Problem Description Given two positive integers G and L, could you tell me how many solutions…

一直在想丝帛题要不要贴呢...后来觉得还是贴了吧...反正没人看...blog是开给自己看的...偶尔无聊打打blog也显得生活非常充实... 题意: 给一个gcd和lcm求满足啊他们的最小的a和b. SOL: 还想着质因数分解来着...转念一想比gcd小的数的gcd也不能变成G啊...这不是丝帛么...那最小的不就是gcd么... 数论用几个丝帛题开头好了 = =... Code: #include <cstdio> #include <cstdlib> #include <…

GCD and LCM Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=4497 Description Given two positive integers G and L, could you tell me how many solutions of (x, y, z) there are, satisfying that gcd(x, y, z) = G and…

并不重要的前言 最近学习了一些数论知识,但是自己都不懂自己到底学了些什么qwq,在这里把知识一并总结起来. 也不是很难的gcd和lcm 显而易见的结论: 为什么呢? 根据唯一分解定理: a和b都可被分解为素因子的乘积,形如: 则显而易见的有一下结论: 相乘,得: 得证 几种求gcd的算法 欧几里得算法(辗转相除法) 辗转相减法(优化:stein_gcd) 欧几里得算法 基于事实: 实现: int gcd(int a, int b){ ) ? a : gcd( b , a % b) ; } 简短而…

GCD & LCM Inverse Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9928   Accepted: 1843 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 a…

GCD and LCM Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)Total Submission(s): 40    Accepted Submission(s): 22 Problem Description Given two positive integers G and L, could you tell me how many solutions of (x,…