思路：这里就要看往那边贪心了，因为解决的是最大值最小化，最小值最大化。也就是说当满足大于等于c时，l=mid+1这样的二分得到的就是在所有满足条件函数下的最右端.

 

#include
     
      #include
      
       using namespace std;#define ll long longconst int maxn = 1e5 + 10;int a[maxn], n, maxx, ans;ll c, mid;bool check(ll x){    ll sum = 0;    for (int i = 1; i <= n; ++i)        sum += a[i] / x;    return sum >= c;}void half(){    ll l = 1, r = maxx;    while (l <= r){        mid = (r + l) >> 1;        if (check(mid)){ l = mid + 1; }        else r = mid - 1;    }    ans = r;}int main(){    cin >> n >> c;    for (int i = 1; i <= n; ++i)    {        cin >> a[i]; maxx = max(maxx, a[i]);     }    half();        //二分    cout << ans << endl;}