给你N个数求其最小公倍数（hdu 1019 Least Common Multiple）

Problem DescriptionThe 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 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.OutputFor each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.Sample Input23 5 7 156 4 10296 936 1287 792 1Sample Output10510296

题意就是给你T组，每组n个数，求其最小公倍数
思路 可以用求出的两个数值间的最小公倍数和下一个数进行gcd如果是同一个数的话则ans不被覆盖 否则求出新的最小公倍数覆盖ans 不断进行此操作

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>using namespace std;const int N=1e5;long long a[N];long long gcd(long long aa,long long bb){    if(bb==0)        return aa;    else        return gcd(bb,aa%bb);}int main(){    int T;    cin>>T;    while(T--)    {        int n;        cin>>n;        for(int i=1;i<=n;i++)            cin>>a[i];        long long ans=a[1];        for(int i=2;i<=n;i++)            ans=ans/gcd(ans,a[i])*a[i];        cout<<ans<<endl;    }    return 0;}