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…

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…

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…

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…

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…

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…

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…