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…

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…

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--)…

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…

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: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…

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…