solution

其实就是要求一个最大的x使得。

显然当x增大时答案会变小，具有单调性，所以考虑二分答案。

二分一个x，检查答案是否大于等于K即可。

code

/*
* @Author: wxyww
* @Date:   2020-04-16 11:12:25
* @Last Modified time: 2020-04-16 11:15:09
*/
#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<ctime>
using namespace std;
typedef long long ll;
const int N = 200010;
ll read() {
    ll x = 0,f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1; c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = x * 10 + c - '0'; c = getchar();
    }
    return x * f;
}
int n,a[N];
int check(int x) {
    int ret = 0;
    for(int i = 1;i <= n;++i) ret += a[i] / x;
    return ret;
}
int main() {
    n = read();int K = read();
    for(int i = 1;i <= n;++i) a[i] = read();
    int l = 1,ans = 0,r = 1e9;
    while(l <= r) {
        int mid = (l + r) >> 1;
        if(check(mid) >= K) ans = mid,l = mid + 1;
        else r = mid - 1;
    }
    cout<<ans;
    return 0;
}