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--)…

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…

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…

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…

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…

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.  …

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…