HDU 1019 Least Common Multiple

求多个数的lcm,水题~

int n;

int gcd(int a,int b)
{
    return b?gcd(b,a%b):a;
}

int lcm(int a,int b)
{
    return a/gcd(a,b)*b;
}

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        cin>>n;
        vector<int> v(n);

        for(int i=0;i<n;i++) cin>>v[i];

        for(int i=0;i<n-1;i++)
            v[i+1]=lcm(v[i],v[i+1]);
        cout<<v[n-1]<<endl;
    }

    //system("pause");
    return 0;
}
posted @ 2021-02-12 12:21  Dazzling!  阅读(29)  评论(0编辑  收藏  举报