結果
| 問題 |
No.3126 Dual Query Problem
|
| コンテスト | |
| ユーザー |
👑 |
| 提出日時 | 2025-04-25 21:36:05 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 2,746 bytes |
| コンパイル時間 | 4,511 ms |
| コンパイル使用メモリ | 262,488 KB |
| 実行使用メモリ | 11,092 KB |
| 最終ジャッジ日時 | 2025-04-25 21:37:10 |
| 合計ジャッジ時間 | 16,886 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
(要ログイン)
| ファイルパターン | 結果 |
|---|---|
| sample | AC * 2 WA * 1 |
| other | AC * 14 WA * 18 |
ソースコード
#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";
}
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;
}