結果

問題 No.3051 Make All Divisible
ユーザー 👑 binap
提出日時 2025-03-07 23:14:51
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 183 ms / 2,000 ms
コード長 3,174 bytes
コンパイル時間 4,061 ms
コンパイル使用メモリ 255,676 KB
実行使用メモリ 7,844 KB
最終ジャッジ日時 2025-06-20 02:25:48
合計ジャッジ時間 5,668 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
other AC * 32
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;

template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}

template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}

int solve(){
	int n, k;
	cin >> n >> k;
	vector<long long> a(n);
	cin >> a;
	vector<long long> b(n);
	long long MAX = 0;
	long long sum = 0;
	rep(i, n){
		b[i] = a[i] - a[i] / k * k;
		sum += b[i];
		chmax(MAX, b[i]);
	}
	if(sum % k != 0){
		cout << "-1\n";
		return 0;
	}
	using P = pair<long long, long long>;
	priority_queue<P, vector<P>, greater<P>> pq;
	rep(i, n){
		if(a[i] - b[i] >= k) pq.emplace(b[i], i);
	}
//	cout << "?";
//	cout << a;
//	cout << b;
	while(k < pq.size() and MAX > sum / k){
		auto [dummy, i] = pq.top(); pq.pop();
		b[i] += k;
		chmax(MAX, b[i]);
		sum += k;
		if(a[i] - b[i] >= k) pq.emplace(b[i], i);
	}
	while(pq.size() and MAX > sum / k){
		auto [dummy, i] = pq.top(); pq.pop();
		b[i] += k;
		chmax(MAX, b[i]);
		sum += k;
	}
//	cout << b;
	if(MAX <= sum / k){
		cout << sum / k << "\n";
	}else{
		cout << "-1\n";
	}
	return 0;
}

int main(){
	int t;
	cin >> t;
	rep(_, t) solve();
	return 0;
}
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