The primitive way is not that longer

int main() { int n; cin >> n; int cpos = 0, cmin = 0, czero = 0; for (int i = 0; i < n; i++) { int num; cin >> num; if (num < 0) cmin++; else if (num > 0) cpos++; else czero++; } cout << static_cast (cpos) / n << endl << static_cast (cmin) / n << endl << static_cast (czero) / n << endl; return 0; }