F L1-6 分鸽子

题目地址：

https://ac.nowcoder.com/acm/contest/5587/F

基本思路：

二分能每人能分到的鸽子肉数量，然后每次check一下这种情况下能分给的人数是不是大于等于m就行了。

参考代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false)
#define int long long
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define INF 0x3f3f3f3f

inline int read() {
  int x = 0, neg = 1; char op = getchar();
  while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
  while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) { putchar('-'); x = -x; }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}

const int maxn = 1e5 + 10;
int n,m,a[maxn];
bool check(int s){
  int cnt = 0;
  rep(i,1,n){
    cnt += a[i]/s;
  }
  return cnt >= m;
}
signed main() {
  IO;
  cin >> n >> m;
  rep(i,1,n) cin >> a[i];
  int l = 1,r = 1e9+10,ans = 0;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(mid)) ans = mid, l = mid + 1;
    else r = mid - 1;
  }
  cout << ans << '\n';
  return 0;
}