解體心得：
1、一開始我的演算法是找出最大的那個數，再將那個數一倍一倍的相加，但是會超時，因為題目的限制是32bits。（過於天真）
2、運用小學奧數的演算法，多個數的最小公倍數，先將兩個數的最小公倍數求出，再與後面的數求最小公倍數。兩個數的最小公倍數，可以先將兩個數相乘再除兩個數的最小公因數。為了避免溢位，可以先相除再相乘。
3、有一個比較簡單的求最大公因數的庫函式呼叫（__gcd(int a,int b)）,但不是標準庫，標頭檔案使用萬能標頭檔案。

題目：

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 Northmerica 2003, Practice 
 

#include<bits/stdc++.h>
using namespace std;
//為了防止溢位全開long long，gcd函式是找出兩個數的最大公因數（輾轉相除法）；
long long gcd(long long a,long long k)
{
    if(k == 0)
        return a;
    else
        return gcd(k,a%k);
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        long
 
 long k,cnt,a;
        a = cnt = 1;
        while(n--)
        {
            cin>>k;//一邊輸入一邊計算，節省時間
            cnt = a/gcd(a,k)*k;//先除後乘防止溢位；
            a = cnt;
        }
        cout<<cnt<<endl;
    }
}

求最大公因數的庫函式使用：

#include<bits/stdc++.h>
//好像只能使用這個標頭檔案
using namespace std
 
;
int main()
{
    int a,b;
    while(scanf("%d%d",&a,&b)!=EOF)
    {
        printf("%d\n",__gcd(a,b));
//不是標準庫中的函式
    }
}