[Codeforces] Round #648 (Div. 2) C. Rotation Matching

Codeforces Round #648 (Div. 2) C. Rotation Matching

  • Time : O(n)
  • Space : O(n)
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#include <bits/stdc++.h>

#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")//Comment optimisations for interactive problems (use endl)
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")

struct PairHash {
inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }
};

#define MAX_N 1010101
#define INF 987654321
#define EPS 1e-9
#define ll long long
#define pll pair<ll, ll>
#define vpll vector<pll>
#define vall3 vector<array<ll,3>>
#define all3 array<ll,3>
#define all5 array<ll,5>
#define vall5 vector<all5>
#define vll vector<ll>
#define vi vector<int>
#define vs vector<string>
#define usll unordered_set<ll>
#define uspll unordered_set<pll, PairHash>
#define umll unordered_map<ll,ll>
#define vvs vector<vs>
#define vvll vector<vll>
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)

ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }

using namespace std;

ll solve(vll& A, vll& B) {
umll mp;
ll n = A.size(), res = 0;
vll rA(n + 1);
for(ll i = 0; i < n; i++) {
rA[A[i]] = i;
}
for(ll i = 0; i < n; i++)
res = max(res, ++mp[(i - rA[B[i]] + n) % n]);
return res;
}

int main() {
ios_base::sync_with_stdio(0);
ios::sync_with_stdio(0);
cin.tie(0);
ll tc = 1;
//cin>>tc;
for (ll i = 1; i <= tc; i++) {
ll n;
cin>>n;
vll A(n), B(n);
for(ll j = 0; j < n; j++) cin>>A[j];
for(ll j = 0; j < n; j++) cin>>B[j];
cout<<solve(A,B)<<endl;
}
return 0;
}
Author: Song Hayoung
Link: https://songhayoung.github.io/2022/06/22/PS/Codeforces/div2-648-c/
Copyright Notice: All articles in this blog are licensed under CC BY-NC-SA 4.0 unless stating additionally.