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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define mod 1000000007
#define mod1 998244353
#define lli long long int
#define plli pair<lli, lli>
#define REP(i, a, b) for (lli i = a; i < b; i++)
#define REPI(i, a, b) for (lli i = b - 1; i >= a; i--)
#define F first
#define ldb long double
#define deci(x) fixed << setprecision(x)
#define S second
#define PB push_back
#define DB pop_back
#define MP make_pair
#define MT make_tuple
#define G(a, b) get<a>(b)
typedef vector<plli> vii; // three data type shortcuts. They may look cryptic
typedef vector<lli> vi;

#define o_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define o_setlli tree<lli, null_type, less<lli>, rb_tree_tag, tree_order_statistics_node_update>
#define o_multiset tree<int, null_type, std::less_equal<int>, rb_tree_tag, tree_order_statistics_node_update>
#define o_multisetlli tree<plli, null_type, std::less_equal<plli>, rb_tree_tag, tree_order_statistics_node_update>
//member functions :
//1. order_of_key(k) : number of elements strictly lesser than k
//2. find_by_order(k) : k-th element in the set

lli modI(lli a, lli m);
lli gcd(lli a, lli b);
lli powM(lli x, unsigned lli y, unsigned lli m);
void pairsort(int a[], int b[], int n);
void pairsortlli(lli a[], lli b[], lli n);
lli logint(lli x, lli y);
void Miden(lli **p1, lli n);
void Mmult(lli **p1, lli **p2, lli **ans, lli x, lli y, lli z, lli m);
void Mpow(lli **p1, lli **ans, lli n, lli y, lli m);
plli Egcd(lli x, lli y);
#define LSOne(S) (S & (-S))
#define kira ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)

class FenwickTree
{
private:
    vector<lli> ft;

public:
    FenwickTree() {}
    // initialization: n + 1 zeroes, ignore index 0
    FenwickTree(lli n) { ft.assign(n + 1, 0); }

    lli rsq(lli b)
    { // returns RSQ(1, b)
        lli sum = 0;
        for (; b; b -= LSOne(b))
            sum += ft[b];
        return sum;
    }

    lli rsq(lli a, lli b)
    { // returns RSQ(a, b)
        return rsq(b) - (a == 1 ? 0 : rsq(a - 1));
    }

    // adjusts value of the k-th element by v (v can be +ve/inc or -ve/dec)
    void adjust(lli k, lli v)
    { // note: n = ft.size() - 1
        for (; k < (lli)ft.size(); k += LSOne(k))
            ft[k] += v;
    }
};

class UnionFind
{ // OOP style
public:
    vector<lli> p, rank, setSize; // remember: vi is vector<int>
    lli numSets;
    UnionFind(lli N)
    {
        setSize.assign(N, 1);
        numSets = N;
        rank.assign(N, 0);
        p.assign(N, 0);
        for (lli i = 0; i < N; i++)
            p[i] = i;
    }
    lli findSet(lli i) { return (p[i] == i) ? i : (p[i] = findSet(p[i])); }
    bool isSameSet(lli i, lli j) { return findSet(i) == findSet(j); }
    void unionSet(lli i, lli j)
    {
        if (!isSameSet(i, j))
        {
            numSets--;
            lli x = findSet(i), y = findSet(j);
            // rank is used to keep the tree short
            if (rank[x] > rank[y])
            {
                p[y] = x;
                setSize[x] += setSize[y];
            }
            else
            {
                p[x] = y;
                setSize[y] += setSize[x];
                if (rank[x] == rank[y])
                    rank[y]++;
            }
        }
    }
    lli numDisjointSets() { return numSets; }
    lli sizeOfSet(lli i) { return setSize[findSet(i)]; }
};

lli gcd(lli x, lli y)
{
    if (x == 0)
        return y;
    return gcd(y % x, x);
}

plli Egcd(lli x, lli y)
{
    if (x == 0)
        return MP(0, 1);
    plli t = Egcd(y % x, x);
    return MP(t.S - t.F * (y / x), t.F);
}

lli power(lli x, lli y,lli p)
{
    lli res = 1;
    x = x %p;
    while (y > 0)
    {
        if (y & 1)
            res = (res * x)%p;
        //y must be even now
        y = y >> 1; //y=y/2
        x = (x * x) %p;
    }
    return res;
}

lli modI(lli a, lli m)
{
    lli m0 = m, y = 0, x = 1;
    if (m == 1)
        return 0;
    while (a > 1)
    {
        lli q = a / m;
        lli t = m;
        m = a % m;
        a = t;
        t = y;
        y = x - q * y;
        x = t;
    }
    if (x < 0)
        x += m0;
    return x;
}

void pairsort(int a[], int b[], int n)
{
    pair<int, int> v[n];
    REP(i, 0, n)
    {
        v[i].F = a[i];
        v[i].S = b[i];
    }
    sort(v, v + n);
    REP(i, 0, n)
    {
        a[i] = v[i].F;
        b[i] = v[i].S;
    }
}

void pairsortlli(lli a[], lli b[], lli n)
{
    pair<lli, lli> v[n];
    REP(i, 0, n)
    {
        v[i].F = a[i];
        v[i].S = b[i];
    }
    sort(v, v + n);
    REP(i, 0, n)
    {
        a[i] = v[i].F;
        b[i] = v[i].S;
    }
}

lli logint(lli x, lli y)
{
    lli ans = 0;
    lli a = 1;
    for (lli i = 0; i <= x; i++)
    {
        if (x < a)
        {
            return ans;
        }
        ans++;
        a *= y;
    }
    return -1;
}

void Miden(lli **p1, lli n)
{
    lli(*x)[n] = (lli(*)[n])p1;
    REP(i, 0, n)
    {
        REP(j, 0, n)
        {
            x[i][j] = 0;
        }
        x[i][i] = 1;
    }
    return;
}

void Mmult(lli **p1, lli **p2, lli **ans, lli x, lli y, lli z, lli m)
{
    lli(*a)[y] = (lli(*)[y])p1;
    lli(*b)[z] = (lli(*)[z])p2;
    lli(*c)[z] = (lli(*)[z])ans;
    REP(i, 0, x)
    {
        REP(j, 0, z)
        {
            c[i][j] = 0;
            REP(k, 0, y)
            {
                c[i][j] += a[i][k] * b[k][j];
                c[i][j] %= m;
            }
        }
    }
    return;
}

void Mpow(lli **p1, lli **ans, lli n, lli y, lli m)
{
    if (y == 0)
    {
        Miden(ans, n);
        return;
    }
    lli t[n][n];
    Mpow(p1, (lli **)t, n, y / 2, m);
    lli z[n][n];
    Mmult((lli **)t, (lli **)t, (lli **)z, n, n, n, m);
    if (y % 2)
    {
        Mmult((lli **)z, p1, ans, n, n, n, m);
    }
    else
    {
        Miden((lli **)t, n);
        Mmult((lli **)z, (lli **)t, ans, n, n, n, m);
    }
    return;
}

/* comparator function in set */
/*
struct cmp {
	bool operator() (const pair<int, int> &a, const pair<int, int> &b) const {
		int lena = a.second - a.first + 1;
		int lenb = b.second - b.first + 1;
		if (lena == lenb) return a.first < b.first;
		return lena > lenb;
	}
};
*/
/* set<pair<lli,lli>,cmp> s; */  

int main()
{
    /*ifstream stdin;
    stdin.open("abc.txt");
    ofstream stdout;
    stdout.open("sorted.out");*/
    /*freopen("test.txt","r",stdin);
    freopen("output.txt","w",stdout);*/
    kira;
    lli t;
    t=1;
    //cin>>t;
    while(t--)
    {
       lli n,k;
       cin>>n>>k;
       lli a[n],p=0;
       REP(i,0,n)
       {
           cin>>a[i];
           p=max(p,a[i]);
       }
       lli l=1,r=p,ans=p;
       while(l<=r)
       {
           lli mid=l+(r-l)/2,a1=0;
           for(lli i=0;i<n;i++)
           {
               a1+=a[i]/mid;
           }
           //cout<<mid<<" "<<l<<" "<<r<<" "<<a1<<'\n';
           if(a1<=k)
           {
               ans=mid;
               r=mid-1;
           }
           else
           {
               l=mid+1;
           }
       }
       cout<<ans<<'\n';
    }
}