HDU 1019 Least Common Multiple【gcd+lcm+水+多个数的lcm】

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 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 first line of the input
will contain a single integer indicating the number of problem
instances. Each instance will consist of a single line of the form m n1
n2 n3 ... nm where m is the number of integers in the set and n1 ... nm
are the integers. All integers will be positive and lie within the range
of a 32-bit integer.

 

Output

For
each problem instance, output a single line containing the
corresponding LCM. All results will lie in the range of a 32-bit
integer.

 

Sample Input

2
3 5 7 15
6 4 10296 936 1287 792 1

Sample Output

105
10296

Source

East Central North America 2003, Practice

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1019

分析：多个数的最小公倍数，直接每次对两个进行lcm操作就好了，具体代码如下：

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 inline ll gcd(ll a,ll b)
 {
     return b==?a:gcd(b,a%b);
 }
 inline ll lcm(ll a,ll b)
 {
     return a*b/gcd(a,b);
 }
 ll a[];
 int main()
 {
     ll n;
     while(scanf("%lld",&n)!=EOF)
     {
         while(n--)
         {
             ll m;
             scanf("%lld",&m);
             for(ll i=;i<=m;i++)
                 scanf("%lld",&a[i]);
             for(ll i=;i<=m-;i++)
                 a[i+]=lcm(a[i],a[i+]);
             printf("%lld\n",a[m]);
         }
     }
     return ;
 }