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;

// #define ONLINE_JUDGE

#define V vector
#define P pair

using LL = long long;
#define FOR(i, x, y) \
    for (decay<decltype(y)>::type i = (x), _##i = (y); i < _##i; ++i)
#define FORD(i, x, y) \
    for (decay<decltype(x)>::type i = (x), _##i = (y); i > _##i; --i)
#ifndef ONLINE_JUDGE
#define dbg(x...)                             \
    do {                                      \
        cerr << "\033[32;1m" << #x << " -> "; \
        err(x);                               \
    } while (0)
void err() { cerr << "\033[39;0m" << endl; }
template <template <typename...> class T, typename t, typename... A>
void err(T<t> a, A... x) {
    for (auto v : a) cerr << v << ' ';
    err(x...);
}
template <typename T, typename... A>
void err(T a, A... x) {
    cerr << a << ' ';
    err(x...);
}
#else
#define dbg(...)
#endif
#define sz(x) (int)x.size()

LL bin(LL a, LL b, LL p) {
    LL res = 1;
    a = (a % p + p) % p;
    for (; b; b >>= 1, a = a * a % p)
        if (b & 1) res = res * a % p;

    return res;
}

const int N = 2e5+50;
int n, k, a[N];

int check(int m) {
    int cnt = 0;
    for (int i=0; i<n; ++i) {
        cnt += (a[i] - 1) / m;
    }
    if (cnt > k)
        return -1;
    if (cnt+n <= k)
        return 1;
    int cnt2 = 0;
    for (int i=0; i<n; ++i) {
        cnt2 += a[i] % m != 0;
    }
    return cnt2 + cnt <= k;
}

void solve() {
    scanf("%d%d", &n, &k);
    for (int i=0; i<n; ++i)
        scanf("%d", &a[i]);

    int l = 0, r = 1000000000;
    while (l < r-1) {
        int m = (l+r) >> 1;
        
        if (check(m) >= 0) {
            r = m;
        } else {
            l = m;
        }
    }
    printf("%d\n", r + check(r));
}

int main() {

    // int t;
    // cin >> t;
    // while (t--) solve();

    solve();

    return 0;
}