C. Cards Partition
time limit per test: 2 seconds
memory limit per test: 256 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≤10^4). The description of the test cases follows.
The first line of each test case contains two integers n, k (1≤n≤2⋅10^5, 0≤k≤10^16) --- 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≤10^10, ∑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⋅10^5.
Output
For each test case, output a single integer: the maximum possible size of a deck if you operate optimally.
Example
Input
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
Output
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.
【思路分析】
思维。显然划分后每组牌数（即size）不超过n，那么要求size最大，只需要从n到1枚举size即可。对每个size，计算出还需要购买的牌数，跟k作比较即可。
cpp
#include<bits/stdc++.h>
#define i64 long long
using namespace std;
void solve() {
i64 n, k;
cin >> n >> k;
i64 maxn = 0, sum = 0;
i64 a[n];
for (int i = 0; i < n; i++) {
cin >> a[i];
maxn = max(maxn, a[i]);
sum += a[i];
}
while (n >= 0) {
i64 cnt = max(maxn, (sum + n - 1) / n);
if (cnt * n - sum <= k) break;
n--;
}
cout << n << endl;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
int t = 1;
cin >> t;
while (t--) {
solve();
}
return 0;
}