/* There are a number of people who will be attending ACM-ICPC World Finals. Each of them may be well versed in a number of topics. Given a list of topics known by each attendee, you must determine the maximum number of topics a 2-person team can know. Also find out how many ways a team can be formed to know that many topics. Lists will be in the form of bit strings, where eachstring represents an attendee and each position in that string represents a field of knowledge, 1 if its a known field or 0 if not. Ipnut : The first line contains two space-separated integers and , where represents the number of attendees and represents the number of topics. Each of the next lines contains a binary string of length . If the th line's th character is , then the th person knows the topic. example 4 5 10101 11100 11010 00101 Ouput : On the first line, print the maximum number of topics a 2-person team can know. On the second line, print the number of ways to form a 2-person team that knows the maximum number of topics. example from the above input : 5 2 explanation : Calculating topics known for all permutations of 2 attendees we get: (1,2) -> 4 (1,3) -> 5 (1,4) -> 3 (2,3) -> 4 (2,4) -> 4 (2,4) -> 5 The 2 teams (1, 3) and (3, 4) know all 5 topics which is maximal. */ #include using namespace std; vector split_string(string); // Complete the acmTeam function below. vector acmTeam(vector topic) { int n = topic[0].length(); int m = topic.size(); vector ans; int max =0; int it=0; for (int i = 0; i < n-1; i++) { for (int k = i+1; k < n; k++) { for(int j=0;jmax) { it=max; } } } ans.push_back(n); ans.push_back(max); return ans; } int main() { ofstream fout(getenv("OUTPUT_PATH")); string nm_temp; getline(cin, nm_temp); vector nm = split_string(nm_temp); int n = stoi(nm[0]); int m = stoi(nm[1]); vector topic(n); for (int i = 0; i < n; i++) { string topic_item; getline(cin, topic_item); topic[i] = topic_item; } vector result = acmTeam(topic); for (int i = 0; i < result.size(); i++) { fout << result[i]; if (i != result.size() - 1) { fout << "\n"; } } fout << "\n"; fout.close(); return 0; } vector split_string(string input_string) { string::iterator new_end = unique(input_string.begin(), input_string.end(), [] (const char &x, const char &y) { return x == y and x == ' '; }); input_string.erase(new_end, input_string.end()); while (input_string[input_string.length() - 1] == ' ') { input_string.pop_back(); } vector splits; char delimiter = ' '; size_t i = 0; size_t pos = input_string.find(delimiter); while (pos != string::npos) { splits.push_back(input_string.substr(i, pos - i)); i = pos + 1; pos = input_string.find(delimiter, i); } splits.push_back(input_string.substr(i, min(pos, input_string.length()) - i + 1)); return splits; }

ACM_PC team