// Coded by Santanu Sikder /* Little explanation about this program: You will be prompted to input the number of teams (N) you desire to make a fixture for. Note that THIS PROGRAM SHOWS "TIME LIMIT EXCEEDED" IF N > 5 ON SOLOLEARN COMPILER! Then you must write the name of each team on a new line (by presing enter). The cyclic method will organise the teams into two columns in each round and the pair of teams in the same row will compete against each other in that round. For odd number of teams, we are going to have one 'Bye'(a concept used to adjust fixtures in such cases; it is simply a privilege given to one team in a particular round such that the team isn't required to play that round) which will correspond to the team beside the Bye (in the same row). Therefore, there will be N rounds. In the case of even number of teams, there won't be any bye and total number of rounds will equal to (N-1). */ #include using namespace std; int main() { int N = 0, R = 0, rows = 0, size = 0; cout << "Please enter the number of teams, N: "; cin >> N; cout << N; string teams[N]; ((N % 2) == 0) ? (size = N) : (size = N + 1); string teamsF[size]; if(N < 2) { cout << "\n\nSorry, number of teams should be ≥ 2."; } else { for(int x = 0; x < N; x++) { cin >> teams[x]; if(teams[x] == "") { cout << "\n\nPlease enter valid input!"; } } } if((N % 2) == 0) { R = N - 1; rows = N / 2; for(int h = 0; h < N; h++) { teamsF[h] = teams[h]; } } else { R = N; rows = (N + 1) / 2; for(int g = 0; g <= N; g++) { if(g == N) { teamsF[g] = "(Bye)"; } else { teamsF[g] = teams[g]; } } } cout << "\nSo the required fixture is:" << endl; for(int r = 0; r < R; r++) { for(int s = 0; s < rows; s++) { int dif = s - r, add = s + r, ind = 0, ind2 = 0; ind = dif; ind2 = R - add; if(ind < 0) { ind = R + ind; } if(ind2 < 0) { ind2 = R + ind2; } if(s == 0) { cout << endl << teamsF[ind] << " - " << teamsF[R]; } else { cout << endl << teamsF[ind] << " - " << teamsF[ind2]; } } cout << endl; } return 0; }

League Fixture using Cyclic Method