結果

問題 No.3126 Dual Query Problem
ユーザー 👑 binap
提出日時 2025-04-25 21:36:23
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 128 ms / 2,000 ms
コード長 2,758 bytes
コンパイル時間 4,544 ms
コンパイル使用メモリ 261,740 KB
実行使用メモリ 11,044 KB
最終ジャッジ日時 2025-06-20 02:41:20
合計ジャッジ時間 13,825 ms
ジャッジサーバーID
(参考情報)
judge5 / judge3
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
sample AC * 3
other AC * 33
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 pair<int, int> P;

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 main(){
	int n, q;
	cin >> n >> q;
	vector<int> a(n);
	cin >> a;
	map<int, int> ma;
	int cnt = 1;
	
	vector<tuple<int, int, int>> Query;
	rep(i, n){
		if(ma.find(a[i]) == ma.end()){
			Query.emplace_back(1, cnt, a[i]);
			Query.emplace_back(2, cnt, 0);
			ma[a[i]] = cnt;
			cnt++;
		}else{
			Query.emplace_back(2, ma[a[i]], 0);
		}
	}
	if(Query.size() > q){
		cout << "No\n";
		return 0;
	}
	while(Query.size() < q){
		Query.emplace_back(1, 1, 1);
	}
	cout << "Yes\n";
	for(auto [t, var1, var2] : Query){
		if(t == 1) cout << 1 << ' ' << var1 << ' ' << var2 << "\n";
		if(t == 2) cout << 2 << ' ' << var1 << "\n";
	}
	return 0;
}
0