Codeforces · 2193B

Reverse a Permutation

This C++ solution uses mathematical reasoning for Codeforces 2193B Reverse a Permutation. Read the reasoning, inspect the code, or try your own test case below.

2193BMathMathematical reasoningC++43 lines
Solution175of 248
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Approach

Mathematical reasoning

Reverse a Permutation matches the one-reversal objective and lexicographically maximum construction.

Math

Problem and code

Useful links.

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

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Implementation

B_Reverse_a_Permutation.cpp

C++

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    const int inf = 1e9;
    const long long INF = 1e17; //❄️
    void solve() {
        int n; cin >> n;
        vector<int> p(n);
        for (int i = 0; i < n; i++) {
            cin >> p[i];
        }
        int end = -1;
        for (int i = 0; i < n; i++) {
            if (p[i] != n - i) {
                end = i; break;
            }
        }
        if (end == -1) {
            for (int i = 0; i < n; i++) {
                cout << n - i << " \n"[i == n - 1];
            }
            return;
        }
        int start = find(p.begin(), p.end(), n - end) - p.begin();
        for (int i = 0; i < end; i++) {
            cout << p[i] << " ";
        }
        for (int i = start; i >= end; i--) {
            cout << p[i] << " ";
        }
        for (int i = start + 1; i < n; i++) {
            cout << p[i] << " ";
        }
        cout << '\n';
    }
    int main() {
        ios::sync_with_stdio(0); cin.tie(0); 
        int t; cin >> t;
        while (t--) {
            solve();
        }
        return 0;
    }
        

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