結果

問題 No.2254 Reverse Only
ユーザー omgflakaomgflaka
提出日時 2023-03-24 21:51:04
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 3,048 bytes
コンパイル時間 2,320 ms
コンパイル使用メモリ 209,720 KB
実行使用メモリ 6,416 KB
最終ジャッジ日時 2023-10-18 20:59:11
合計ジャッジ時間 5,740 ms
ジャッジサーバーID
（参考情報） 		judge12 / judge14
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テストケース

テストケース表示
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ソースコード

diff # 

#include <bits/stdc++.h>
#pragma GCC optimize("O3,Ofast,unroll-loops")
#pragma GCC target("avx2,sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
typedef long long           ll;
#define vi                  vector<int> 
#define vll                 vector<ll> 
#define vllp                vector<pair<ll,ll>>> 
#define vb                  vector<bool> 
#define pub                 push_back
#define emb                 emplace_back
#define rep(i,a,b)          for(long long i = a; i < b; i++)
#define rrep(i,a,b)         for(long long i = a; i >= b; i--)
#define iterm(key,val,ok)   for(auto const& [key, val]:ok)
#define all(x)              x.begin(),x.end()
#define py                  cout<<"YES"<<endl
#define pn                  cout<<"NO"<<endl
#define pa                  cout<<ans<<endl
#define pe                  cout<<endl
#define mp                  make_pair
#define f                   first 
#define se                  second 
using pll=pair<ll,ll>;
template<typename T>        istream& operator>>(istream& is, vector<T>& v) { for (auto& i : v) is >> i;        return is; }
template<typename T>        ostream& operator<<(ostream& os, const vector<T>& v) { for (auto& i : v) os << i << " "; return os; }
template<typename T>        istream& operator>>(istream& is, pair<T, T>& v) { is >> v.first >> v.second;        return is; }
template<typename T>        ostream& operator<<(ostream& os, const pair<T, T>& v) { os << v.first << " " << v.second; return os; }
 
 
 
/*
problems with mod or small constraints are DP (2d,3d) or combo
ascii: 1=49 a=97 A=65
check edge cases and constraints CHECK FOR JUST ONE ELEMENT OR SAME NUMBERS ETC
always think about reversing a process
see if theres any dumb tricks if youre not getting it: like only having to check small values or sth
(a-b)modm = (a%m+(m-b%m))%m
mark visited after adding to queue
think about transforming representations of graphs, if some graph is too big to be represented, think about bipartite
representation, if its too small maybe you can expand according to problem to make it easier
you can also bfs on pairs of nodes, different representations of nodes etc etc
lower bound can give beyond end also (check segfaults etc if using)
many smol vectors are slowwww, use std:array instead
to search for and remove particular things fastly just use multiset, like u can use prefix multiset and find closest
value of something also like so many applications in just logn time
READ ALL FUCKING SAMPLES, READ EVERYTHING BEFORE PUTTING PEN TO PAPER, START FROM THE SIMPLEST SAMPLE
*/
 
 
void solve(){
    ll n,k;cin>>n>>k;
    vector<ll> a(n),b(n);cin>>a>>b;
    if(a==b){
        py;return;
    }
    reverse(all(a));
    if(a==b && k==n){
        py;return;
    }
    if(k>n-1){
        pn;return;
    }
    sort(all(a));sort(all(b));
    if(a==b){
        py;return;
    }
    pn;
    

}
 
signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    int t=1;
    //cin>>t;
    while(t--){
        solve();
    }
}
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