小测D

就是二分查找就够了，找到符合条件的那个最小值
不会二分可以去学一下，可以看看这个：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
",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;
}