Problem:
Score : 500 points

Problem Statement

We have N logs of lengths A_1,A_2,\cdots A_N.

We can cut these logs at most K times in total. When a log of length L is cut at a point whose distance from an end of the log is t (0<t<L), it becomes two logs of lengths t and L-t.

Find the shortest possible length of the longest log after at most K cuts, and print it after rounding up to an integer.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq K \leq 10^9

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \cdots A_N

Output

Print an integer representing the answer.

Sample Input 1

2 3
7 9

Sample Output 1

4

First, we will cut the log of length 7 at a point whose distance from an end of the log is 3.5, resulting in two logs of length 3.5 each.

Next, we will cut the log of length 9 at a point whose distance from an end of the log is 3, resulting in two logs of length 3 and 6.

Lastly, we will cut the log of length 6 at a point whose distance from an end of the log is 3.3, resulting in two logs of length 3.3 and 2.7.

In this case, the longest length of a log will be 3.5, which is the shortest possible result. After rounding up to an integer, the output should be 4.

Sample Input 2

3 0
3 4 5

Sample Output 2

5

Sample Input 3

10 10
158260522 877914575 602436426 24979445 861648772 623690081 433933447 476190629 262703497 211047202

Sample Output 3

292638192

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)


#include<bits/stdc++.h>

#define ll long long
#define ld long double
#define FOR(x,y) for(ll i=x;i<=y;i++)
#define pb(x) push_back(x)
#define mp make_pair
#define pii pair<int,int>
#define pll pair<ll,ll>
#define vii vector<int>
#define vll vector<ll>
#define matrix(x) vector<vector<x>>
#define vss vector<string>
#define PI (2*acos(0.0))
#define sqr(x) ((x)*(x))
#define sf(a)   scanf("%d",&a)
#define sfl(a)  scanf("%lld",&a)
#define sff(a,b)    scanf("%d %d",&a,&b)
#define sffl(a,b)   scanf("%lld %lld",&a,&b)
#define sfff(a,b,c) scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)    scanf("%lld %lld %lld",&a,&b,&c)
#define ms(a,b) memset(a, b, sizeof(a))
#define ff  first
#define ss  second
#define lcm(a, b)   ((a)*((b)/__gcd(a,b)))
#define all(a)  a.begin(),a.end()
#define readfile freopen("input.txt","r",stdin);
#define writefile freopen("output.txt","w",stdout);
#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL);
#define gap " "
#define mx 104
#define inf (ll)1e9
#define WHITE 1
#define GRAY 2
#define BLACK 3
#define EMPTY_VALUE -1
#define mod 1000000007
#define MOD(a,b) (a%b + b)%b

using namespace std;

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

ll x,k;
vll a;

bool possible(ll mid)
{
    if(mid<=0) return false;
    ll cnt = 0;
    for(ll i=0;i<x;i++)
    {
        if(a[i]<=mid) continue;
        cnt += ceil(a[i]/(mid+0.0))-1;
        //cnt += (a[i]-1)/mid;
    }
    return cnt<=k;
}

int main()
{
    //fastio
    sffl(x,k);
    a.resize(x+1);
    for(ll i=0;i<x;i++) sfl(a[i]);
    ll mini = 0,maxi = 1e9+100;
    
    while (mini<maxi)
    {
        ll mid = (mini+maxi)/2;
        //cout<<mid<<endl;
        if(possible(mid))
        {
            maxi = mid;
        }
        else
        {
            mini = mid+1;
        }
    }
    cout<<mini<<endl;
    

    return 0;
}