就是二分查詢就夠了，找到符合條件的那個最小值
不會二分可以去學一下，可以看看這個：https://www.cnblogs.com/wzl19981116/p/9354012.html

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#define sf scanf
#define scf(x) scanf("%d",&x)
#define scff(x,y) scanf("%d%d",&x,&y)
#define scfff(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define vi vector<int>
#define mp make_pair
#define pf printf
#define prf(x) printf("%d\n",x)
#define mm(x,b) memset((x),(b),sizeof(x))
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
int in(){
    int x=0,f=1;  char ch;
    while(ch<'0'||ch>'9')  {if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return f*x;}
using namespace std;
const ll mod=1e9+7;
const double eps=1e-6;
const double pi=acos(-1.0);
const int inf=0x7fffffff;
const int N=1e5+7;
ll n,m,a[N];
bool judge(ll t)
{
    ll s=0;
    rep(i,0,n)
    {
        s=s+t/a[i];
    }
    return s>=m;
 } 
int main()
{
    cin>>n>>m;
    rep(i,0,n)
        cin>>a[i];
    ll l=0,r=100000000009;
    while(r-l>1)
    {
        ll mid=(l+r)/2;
        if(judge(mid))
            r=mid;
        else 
            l=mid;
    }
    while(judge(r)) r--;
    r++;
    cout<<r;
    return 0;
}