解题报告:求多个数的最小公倍数,其实还是一样,只需要一个一个求就行了,先将答案初始化为1,然后让这个数依次跟其他的每个数进行求最小公倍数,最后求出来的就是所有的数的最小公倍数.也就是多次GCD． #include<cstdio> #include<iostream> #include<cstring> using namespace std; typedef __int64 INT; INT GCD(INT a,INT b) { ? b:GCD(b,a%b); } in…

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…

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…

http://acm.hdu.edu.cn/showproblem.php?pid=1019 LCM即各数各质因数的最大值,搞个map乱弄一下就可以了. #include<bits/stdc++.h> using namespace std; typedef long long ll; typedef unsigned int ui; map<ui,ui> M; ll _pow(ui f,ui s){ ll res=1; while(s){ res*=f; s--; } retur…

求一组数据的最小公倍数. 先求公约数在求公倍数.利用公倍数,连续求全部数的公倍数就能够了. #include <stdio.h> int GCD(int a, int b) { return b? GCD(b, a%b) : a; } inline int LCM(int a, int b) { return a / GCD(a, b) * b; } int main() { int T, m, a, b; scanf("%d", &T); while (T--)…

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…

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.  …

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…

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…

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…

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…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1019 解题思路:lcm(a,b)=a*b/gcd(a,b) 反思:最开始提交的时候WA,以为是溢出了,于是改成了long long,还是WA,于是就不明白了,于是就去看了discuss,发现应该这样来写 lcm(a,b)=a*gcd(a,b)*b;是为了以防a乘以b太大溢出,注意啊!!!!所以就先除再乘. #include<stdio.h> int gcd(int a,int b) { int t…

http://acm.hdu.edu.cn/showproblem.php?pid=2028 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 为每组测试数据输出它们的最小公倍数,每个测试实例的输出占一行.你可以假设最后的输出是一个32位的整数.   Sample Input 2 4 6 3 2 5 7   Sample Output 12 70   代码: #includ…

题意:求所有自己的最小公倍数的和. 该集合是  2^ai  * 3^bi 思路:线段树. 线段树中存的是  [3^b * f(b)]   f(b)表示 因子3 的最小公倍数3的部分  为 3^b的个数  那么从小到大枚举a  对于当前的  ab  ,  如果之前的b小于当前的b  那么最小公倍数就为  (2^a) *  (3^b)   个数 就为 2^x     x表示a  b 都小于当前a b的个数 .  大于的部分 就直接是  2^a   * 线段树上[b,max]的和.   求好当前更新进…

思路: 容易知道,分解成素数的lcm肯定是最大的,因为假设分解成2个合数,设定x为他们的 最大公约数, 那么他们的最小公倍数就要减少x倍了 然后如果是素数之间的最小公倍数,那么就只是他们的乘积,同样的n分解,没有 除的肯定比有除的大, 因此可以得到结论 所以可以先晒一次素数,然后用这些素数填满那个n 这里填满也很容易想到是背包问题了,因为同一个素数可以用几次,所以就是一个 典型的多重背包了, 就是dp[j] = lcm(dp[j - k] , dp[k]); 然后还有一个问题,就是对于所有素数取…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1019 Least Common Multiple Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 61592    Accepted Submission(s): 23486 Problem Description The least comm…

http://acm.hdu.edu.cn/showproblem.php?pid=1019 Least Common Multiple Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 25035    Accepted Submission(s): 9429 Problem Description The least common m…

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…

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…

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…

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…

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…

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…

***************************************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    …

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",&…

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…

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…

题目地址:http://ac.jobdu.com/problem.php?pid=1056 题目描述: 输入两个正整数,求其最大公约数. 输入: 测试数据有多组,每组输入两个正整数. 输出: 对于每组输入,请输出其最大公约数. 样例输入: 49 14 样例输出: 7 来源: 2011年哈尔滨工业大学计算机研究生机试真题 #include <stdio.h> int gcd1 (int a, int b){ if (b == 0) return a; else return gcd1 (b, a…