C. Cards Partition

思路：

可以O(n)直接判断，牌组从大到小依次遍历即可。
不要用二分答案，因为答案不一定是单调的

代码:

#include <bits/stdc++.h>
#define endl '\n'
#define int long long
#define pb push_back
#define pii pair<int,int>
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
typedef long long ll;
using namespace std;
int a[200005];
int n, k, sum = 0, ans = 0,maxn=0;

bool check(int x){
	if(x==1)return true;
	int mzs=0;
	if(n>1){mzs = (sum-maxn)/(x-1);}
	int zs=sum/x;
	if(zs > mzs) zs = mzs;
	int ys=sum - zs*x;
	int dy=max(0LL,maxn-zs);
	if(ys>0)dy=max(1LL,dy);
	return (dy*x)<=k+ys;
}


void solve() {
	sum = 0;
	ans = 1;
	maxn = 0;
	memset(a, 0, sizeof(a));
	cin >> n >> k;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		maxn = max(maxn,a[i]);
		sum += a[i];
	}

	for (int i = n; i >= 1; i--) {
		if (check(i)) {
			ans = i;
			break;
		}
	}

	cout << ans << endl;
}

signed main() {
	cin.tie(0)->ios::sync_with_stdio(0);
	int T = 1;
	cin >> T;
	while (T--) {
		solve();
	}
	return 0;
}