Problem Description 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 and 15 is 105. Input Input will consist of multiple prob…

太简单了...题目都不想贴了 //算n个数的最小公倍数 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; int gcd(int a, int b) { ?a:gcd(b,a%b); } int lcm(int a, int b) { return a/gcd(a,b)*b; } int main() { int T; scanf("%d",&…

Least Common Multiple Time Limit: 2 Seconds      Memory Limit: 65536 KB 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 and…

  思路: 求第一个和第二个元素的最小公倍数,然后拿求得的最小公倍数和第三个元素求最小公倍数,继续下去,直到没有元素 注意:通过最大公约数求最小公倍数的时候,先除再乘,避免溢出   #include <iostream> #include <cmath> #include <cstdio> #include <vector> #include <string.h> #include <string> #include <algo…

Lowest Common Multiple Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 34980    Accepted Submission(s): 14272 Problem Description 求n个数的最小公倍数.   Input 输入包含多个测试实例,每个测试实例的开始是一个正整数n,然后是n个正整数.  …

也称欧几里得算法 原理: gcd(a,b)=gcd(b,a mod b) 边界条件为 gcd(a,0)=a; 其中mod 为求余 故辗转相除法可简单的表示为: int gcd(int a, int b) { return b ==0? a:gcd( b, a% b); } 简洁而优雅. 例如:HDU 2028 Lowest Common Multiple Plus求n个数的最小公倍数. 最小公倍数=两数之积  /  最大公约数 这里防止中间过程溢出,先除以最大公约数,然后在求积. #includ…

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 and 15 is 105. Input Input will consist of multiple problem instances. The f…

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 and 15 is 105. InputInput will consist of multiple problem instances. The fi…

题目链接:http://ac.jobdu.com/problem.php?pid=1439 详解链接:https://github.com/zpfbuaa/JobduInCPlusPlus 参考代码: // // 1439 Least Common Multiple.cpp // Jobdu // // Created by PengFei_Zheng on 10/04/2017. // Copyright © 2017 PengFei_Zheng. All rights reserved. /…

Least Common Multiple Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 64855 Accepted Submission(s): 24737 Problem Description The least common multiple (LCM) of a set of positive integers is the sm…

Least Common Multiple Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 42735    Accepted Submission(s): 16055 Problem Description The least common multiple (LCM) of a set of positive integers is…

Lowest Common Multiple Plus Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 33474 Accepted Submission(s): 13683 Problem Description 求n个数的最小公倍数. Input 输入包含多个测试实例,每个测试实例的开始是一个正整数n,然后是n个正整数. Output 为…

***************************************转载请注明出处:http://blog.csdn.net/lttree*************************************** Least Common Multiple Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 28975    …

题目:Least common multiple 链接:http://acm.hdu.edu.cn/showproblem.php?pid=4913 题意:有一个集合s,包含x1,x2,...,xn,有xi=2^ai * 3^bi,然后给你a数组和b数组,求s所有子集合的最小公倍数之和.比如S={18,12,18},那么有{18},{12},{18},{18,12},{18,18},{12,18},{18,12,18},所以答案是174. 思路: 1. 最小公倍数,因为xi只包含两个质因子2.3…

Problem Description 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 and 15 is 105.   Input Input will consist of multiple pr…

Description Partychen like to do mathematical problems. One day, when he was doing on a least common multiple(LCM) problem, he suddenly thought of a very interesting question: if given a number of S, and we divided S into some numbers , then what is…

题目 找出能被两个给定参数和它们之间的连续数字整除的最小公倍数. 范围是两个数字构成的数组,两个数字不一定按数字顺序排序. 例如对 1 和 3 —— 找出能被 1 和 3 和它们之间所有数字整除的最小公倍数. 提示 Smallest Common Multiple 测试用例 smallestCommons([1, 5]) 应该返回一个数字. smallestCommons([1, 5]) 应该返回 60. smallestCommons([5, 1]) 应该返回 60. smallestComm…

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description 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. Fo…

Lowest Common Multiple Plus Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 30907    Accepted Submission(s): 12528 Problem Description 求n个数的最小公倍数.   Input 输入包含多个测试实例,每个测试实例的开始是一个正整数n,然后是n个正整数.  …

Problem Description 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 and 15 is 105.   Input Input will consist of multiple pr…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 51959    Accepted Submission(s): 19706   Problem Description The least common multiple (LCM) of a set of positive integers is the smallest positiv…

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…

题目:Lowest common multiple plus 代码: #include<stdio.h> int common(int a,int b)//计算最大公约数 { int c=a%b,t=0; if(b>a) { t=b; b=a; a=t; } while(a%b!=0) { c=a%b; a=b; b=c; } return b; } int q[105]; int main() { int n,i,j,t=0; while(scanf("%d",&a…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=3092 Least common multiple Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others) 问题描述 Partychen like to do mathematical problems. One day, when he was doing on a least common m…

地址:http://acm.hdu.edu.cn/showproblem.php?pid=1019 题目: Problem Description 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 an…

作为一个oier,以及大学acm党背包是必不可少的一部分.好久没做背包类动规了.久违地练习下-.- dd__engi的背包九讲:http://love-oriented.com/pack/ 鸣谢http://blog.csdn.net/eagle_or_snail/article/details/50987044,这里有大部分比较有趣的dp练手题. hud 2602 01背包板子题 #include<cstdio> #include<iostream> #include<cs…

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 and 15 is 105. 输入 Input will consist of multiple problem instances. The…

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…

[九度OJ]题目1439:Least Common Multiple 解题报告 标签(空格分隔): 九度OJ 原题地址:http://ac.jobdu.com/problem.php?pid=1439 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…

Problem Introduction The least common multiple of two positive integers \(a\) and \(b\) is the least positive integer \(m\) that is divisible by both \(a\) and \(b\). Problem Description Task.Given two integers \(a\) and \(b\), find their least commo…