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;
}

标签：HDU,Multiple,gcd,int,cin,1019,return,lcm
来源： https://www.cnblogs.com/fxh0707/p/14398672.html