pls come up with new solution because mine fails on pretest 4 :waaa:
C. Cards Partition
time limit per test2 seconds
memory limit per test256 megabytes
DJ Genki vs Gram - Einherjar Joker
⠀
You have some cards. An integer between 1
 and n
 is written on each card: specifically, for each i
 from 1
 to n
, you have ai
 cards which have the number i
 written on them.

There is also a shop which contains unlimited cards of each type. You have k
 coins, so you can buy at most k
 new cards in total, and the cards you buy can contain any integer between 1
 and n
, inclusive.

After buying the new cards, you must partition all your cards into decks, according to the following rules:

all the decks must have the same size;
there are no pairs of cards with the same value in the same deck.
Find the maximum possible size of a deck after buying cards and partitioning them optimally.

Input
Each test contains multiple test cases. The first line contains the number of test cases t
 (1≤t≤104
). The description of the test cases follows.

The first line of each test case contains two integers n
, k
 (1≤n≤2⋅105
, 0≤k≤1016
) — the number of distinct types of cards and the number of coins.

The second line of each test case contains n
 integers a1,a2,…,an
 (0≤ai≤1010
, ∑ai≥1
) — the number of cards of type i
 you have at the beginning, for each 1≤i≤n
.

It is guaranteed that the sum of n
 over all test cases does not exceed 2⋅105
.

Output
For each test case, output a single integer: the maximum possible size of a deck if you operate optimally.

Example
InputCopy
9
3 1
3 2 2
5 4
2 6 1 2 4
2 100
1410065408 10000000000
10 8
7 4 6 6 9 3 10 2 8 7
2 12
2 2
2 70
0 1
1 0
1
3 0
2 1 2
3 1
0 3 3
OutputCopy
2
3
1
7
2
2
1
1
2
Note
In the first test case, you can buy one card with the number 1
, and your cards become [1,1,1,1,2,2,3,3]
. You can partition them into the decks [1,2],[1,2],[1,3],[1,3]
: they all have size 2
, and they all contain distinct values. You can show that you cannot get a partition with decks of size greater than 2
, so the answer is 2
.

In the second test case, you can buy two cards with the number 1
 and one card with the number 3
, and your cards become [1,1,1,1,2,2,2,2,2,2,3,3,4,4,5,5,5,5]
, which can be partitioned into [1,2,3],[1,2,4],[1,2,5],[1,2,5],[2,3,5],[2,4,5]
. You can show that you cannot get a partition with decks of size greater than 3
, so the answer is 3
.


my current solution:
#include <bits/stdc++.h>
#define int long long
using namespace std;
 
signed main()
{
    cin.tie(0);
    ios_base::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) {
        int n, k;
        cin >> n >> k;
        vector<int>a(n);
        int sum = 0, maxa = 0;
        for (auto &x : a) {
            cin >> x;
            sum += x;
            if (x > maxa) maxa = x;
        }
        int l = 1, r = n, ans = 1;
        while (l <= r) {
            int mid = l + (r - l) / 2;
            int D1 = (sum + mid - 1) / mid;
            int D = max(D1, maxa);
            if (D * mid - sum <= k) {
                ans = mid;
                l = mid + 1;
            }
            else {
                r = mid - 1;
            }
        }
        cout << ans << "\n";
    }
}