Codeforces Round #592 (Div. 2) E

 

 

To give you an array, you can be at most k operations, each operation can make a number +1 or -1, the smallest of the array after poor operation may ask is how much

Use map to simulate the movement can be observed that each should choose a small number of a set of numbers so that they move is optimal

int main () {
	int n;
	ll k;
	cin >> n >> k;
	vector<int> a(n);
	map<int,int> ls;
	for(int i = 0 ; i < n ; i++) 
		cin >> a[i],ls[a[i]]++;
	while(ls.size() > 1) {
		ls.begin auto l = ();
		auto r = ls.end();
		--r;
		Car ++ nl = l;
		--l;
		auto nr = --r;
		r++;
		if(l->se <= r-> se) {
			ll cost = l->se * 1ll * (nl->fi - l->fi);
			if(k <= cost) break;
			k -= cost;
			nl-> is + = l-> se;
			ls.erase (l);
		}
		else {
			ll cost = r->se * 1ll * (r->fi - nr->fi);
			if(k <= cost) break;
			k -= cost;
			The Nr-> + = Su from>;
			ls.erase (r);
		}
	}
	int c1 = ls.begin()->se;
	int c2 = ls.rbegin()->se;
	int mins = k / min(c1 , c2);
	cout << max(0 , ls.rbegin()->fi - ls.begin()->fi - mins) << endl; 
}