LCM Challenge Time Limit: 2000/1000MS (Java/Others)    Memory Limit: 128000/64000KB (Java/Others) Submit Status Problem Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want…

 LCM Challenge Time Limit:1000MS     Memory Limit:64000KB     64bit IO Format:%lld & %llu Submit Status Practice ACdream 1077 Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I…

[codeforces 235]A. LCM Challenge 试题描述 Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it. But I also don't want to use many numbers, so I'll choose three…

A. LCM Challenge 题目连接: http://www.codeforces.com/contest/235/problem/A Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it. But I also don't wa…

A - LCM Challenge Time Limit: 2000/1000MS (Java/Others)    Memory Limit: 128000/64000KB (Java/Others) Submit Status Problem Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I w…

gcd(最大公因数),lcm(最小公倍数) #include<iostream> using namespace std; int gcd(int a,int b)//辗转相除法(欧几里德算法)求最大公约数 { return b ? gcd(b,a%b) : a; } int lcm(int a,int b) { return a*b/gcd(a,b);//最小公倍数 } int main() { int a,b; while(cin>>a>>b) { cout<…

找到3个不超过n的正整数(可以相同),使得它们的lcm(最小公倍数)最大. Solution 可以做得很优雅吧,但我喜欢(只会)暴力一点 根据质数密度分布性质,最后所取的这三个数一定不会比 \(n\) 小太多(实在不行就取三个质数呀),所以我们钦定一个界,然后暴力枚举取最优即可 #include <bits/stdc++.h> using namespace std; #define int long long int n; signed main() { cin>>n; int…

Problem Description Some days ago, I learned the concept of LCM (least common multiple). I've played with it for several times and I want to make a big number with it. But I also don't want to use many numbers, so I'll choose three positive integers…

GCD(最大公约数) (1)辗转相除法(欧几里得算法)(常用) 将两个数a, b相除,如果余数c不等于0,就把b的值给a,c的值给b,直到c等于0,此时最大公约数就是b (2)更相减损术 将两个书中较大的数a减去较小的数b,如果差c等于0,那么最大公约数为b,如果不等于0,则将b的值给a,c的值给b,继续相减知道差等于0 LCM(最小公倍数) 假设x和y的最大公约数是m,最小公倍数是n,则x*y = m*n…

LCM Extreme Time Limit: 3000ms Memory Limit: 131072KB   This problem will be judged on UVALive. Original ID: 596464-bit integer IO format: %lld      Java class name: Main Find the result of the following code:unsigned long long allPairLcm(int n){ uns…