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)
{
return b==?a:gcd(b,a%b);
}
int lcm(int a, int b)
{
return a/gcd(a,b)*b;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
int tm = ;
int a;
for(int i = ; i < n; i++){
scanf("%d",&a);
tm = lcm(tm,a);
}
printf("%d\n",tm);
}
return ;
}
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Recent Contests

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45603    Accepted Submission(s): 17131

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
 
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