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…
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…
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 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…
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): 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. …
***************************************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 …
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…
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",&…
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…
