long long n, k; cin >> n >> k;
// For Odd Part if(k <= (n + 1) / 2){ cout << k * 2 - 1; } // For Even Part else{ cout << (k - (n + 1) / 2) * 2; }
318/A. Even Odds Solution in C++ Codeforces Round #188(Div.2)
long long n, k; cin >> n >> k;
// For Odd Part if(k <= (n + 1) / 2){ cout << k * 2 - 1; } // For Even Part else{ cout << (k - (n + 1) / 2) * 2; }
Rev. | Язык | Кто | Когда | Δ | Комментарий | |
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en1 | K_Sandip_A | 2022-08-18 05:23:52 | 267 | Initial revision (published) |