Codeforces · 2211A

Antimedian Deletion

This C++ solution uses mathematical reasoning for Codeforces 2211A Antimedian Deletion. Read the reasoning, inspect the code, or try your own test case below.

2211AMathMathematical reasoningC++26 lines
Solution016of 248
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Approach

Mathematical reasoning

Antimedian Deletion output conditions match the code's one-operation answer for N=1 and two-operation construction otherwise.

Math

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

A_Antimedian_Deletion.cpp

C++

    #include <bits/stdc++.h>
    using namespace std;
    using ll = long long;
    using i128 = __int128;
    const int inf = 1e9;
    const ll INF = 1e18; //❄️
    void solve() {
        int n; cin >> n;
        vector<int> v(n);
        for (int i = 0; i < n; i++) cin >> v[i];
        if (n == 1) {
            cout << 1 << '\n';
            return;
        }
        for (int i = 0; i < n; i++) {
            cout << 2 << (i == n - 1 ? '\n' : ' ');
        }
    }
    int main() {
        ios::sync_with_stdio(0); cin.tie(0); 
        int t; cin >> t;
        while (t--) {
            solve();
        }
        return 0;
    }
        

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