[Codeforces] Round #688 (Div. 2) B. Suffix Operations

Codeforces Round #688 (Div. 2) B. Suffix Operations

  • Time : O(n)
  • Space : O(1)
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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 202020
#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)

using namespace std;
ll solve(vll& A) {
ll res = 0, ma = 0, n = A.size();
if(n <= 2) return 0;

for(ll i = 0; i < n - 1; i++) {
res += abs(A[i] - A[i + 1]);
}

for(ll i = 0; i < n - 2; i++) {
ma = max(ma, abs(A[i] - A[i + 1]) + abs(A[i + 1] - A[i + 2]) - abs(A[i] - A[i + 2]));
}
return res - max({ma, abs(A[0] - A[1]), abs(A[n-1] - A[n-2])});
}

int main() {
ios_base::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);
for(ll j = 0; j < n; j++) cin>>A[j];
cout<<solve(A)<<endl;
}
return 0;
}
Author: Song Hayoung
Link: https://songhayoung.github.io/2022/06/18/PS/Codeforces/div2-688-b/
Copyright Notice: All articles in this blog are licensed under CC BY-NC-SA 4.0 unless stating additionally.