DMOJ · dmopc18c4p3

Dr. Henri and Ionization

This C++ solution uses simulation for DMOJ dmopc18c4p3 Dr. Henri and Ionization. Read the reasoning, inspect the code, or try your own test case below.

dmopc18c4p3GeometrySimulationC++29 lines
Solution080of 248
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Approach

Simulation

Dr. Henri and Ionization gives individual and opposite-pair removal costs on a circle; the code's parity/minimum adjustment computes the required minimum total cost.

Geometry

Problem and code

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Written by benbenyaojifen. Try the problem first, then compare your approach with the code.

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Implementation

P_3_Dr_Henri_and_Ionization.cpp

C++

    #include <bits/stdc++.h>
    using namespace std;
    using ll = long long;
    const int inf = 1e9;
    const long long INF = 1e17; //❄️
    int main() {
        ios::sync_with_stdio(0); cin.tie(0); 
        int n; cin >> n;
        vector<int> a(n), b(n);
        for (int i = 0; i < n; i++) cin >> a[i];
        int ans = 0, cnt = 0;
        for (int i = 0; i < n; i++) {
            cin >> b[i];
            if (a[i] < b[i]) ans += a[i];
            else {
                ans += b[i];
                cnt++;
            }
        }
        if (cnt & 1) {
            int mn = inf;
            for (int i = 0; i < n; i++) {
                mn = min(mn, abs(a[i] - b[i]));
            }
            ans += mn;
        }
        cout << ans << '\n';
        return 0;
    }
        

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