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

#include <iostream>
using namespace std;

int zu(int a, int b) {
    int r = 1;
    while (r != 0) {
        r = a%b;
        a = b;
        b = r;
    }
    return a;
}
int main() {
    int n;
    cin >> n;
    int k;
    int a, b;
    while (n--) {
        cin >> k;
        cin >> a ;   
        for (int i = 1; i < k;i++) {
            cin >> b;
            a =(a/zu(a, b))*b;
        }
        cout << a << endl;
    }
    system("pause");
    return 0;
}



#include <iostream>
using namespace std;

int zu(int a, int b) {
    int r = 1;
    while (r != 0) {
        r = a%b;
        a = b;
        b = r;
    }
    return a;
}
int main() {
    int n;
    cin >> n;
    int k;
    int a, b;
    while (n--) {
        cin >> k;
        cin >> a>>b ;   
        a== (a / zu(a, b))*b;
        for (int i = 2; i < k;i++) {
            cin >> b;
            a =(a/zu(a, b))*b;
        }
        cout << a << endl;
    }
    system("pause");
    return 0;
}

为什么第一个能ac，第二个不能ac

这两段代码只有一个地方不一样  
zu(a, b)求出a，b的最大公约数，这个应该没有问题
我知道问题在哪儿了，我把问题理解错了，输入的数可能只有一个，
第二个方法假设开始一定有两个或更多的数求公倍数，但可能只有一个数求公倍数，所以题意理解错误

这个不是最后提交的代码，提交时肯定删了的

哦，下次注意了。