[AtCoder] C - ±1 Operation 1

C - ±1 Operation 1

  • Time : O(logn)
  • 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 vs vector<string>
#define usll unordered_set<ll>
#define uspll unordered_set<pll, PairHash>
#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(ll x, ll a, ll d, ll n) {
if(d < 0) return solve(x, d * (n - 1) + a, -d, n);

ll mi = a, ma = d * (n - 1) + a;
if(mi >= x) return mi - x;
if(ma <= x) return x - ma;
ll l = 0, r = n - 1, lb = ma;
while(l <= r) {
ll m = l + (r - l) / 2;
ll now = a + d * m;
if(now >= x) {
lb = min(lb, now);
r = m - 1;
} else l = m + 1;
}
return min(lb - x, x - lb + d);
}

int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
ll tc = 1;
//cin>>tc;
for(ll i = 1; i <= tc; i++) {
ll x, a, d, n;
cin>>x>>a>>d>>n;
cout<<solve(x,a,d,n)<<endl;
}
return 0;
}
Author: Song Hayoung
Link: https://songhayoung.github.io/2022/06/14/PS/AtCoder/abc255-c/
Copyright Notice: All articles in this blog are licensed under CC BY-NC-SA 4.0 unless stating additionally.