/* * * @ Javier Felipe Toribio 2020-02-14 * * challenge by John Wells * * Graph Factors * * You must read a number from the user. If they * gave you one, it must be between one and * 99,999. If they did not give you any or it was * out of range, use 100. Randomly pick positive * numbers from 1 up to the user's choice. * For each number, count each factor of it in * this list of primes: * 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29. * If none, increment an other count. Stop picking * new numbers once one of the counts reaches * beyond the maximum columns you can display in * portrait to less than the maximum for * landscape. Once complete, graph the counts as * shown. Each asterick is one time being used as * a factor. * * 2 *********************************** * 3 ***************** * 5 ******* * 7 ******* * 11 **** * 13 *** * 17 ****** * 19 ******* * 23 *** * 29 ** * 0 ****** * * https://code.sololearn.com/cB3rFho3Ix4I */ #include #include #include #include #include #include using namespace std; int main() { const int screen_width = 39; vector factors = {2,3,5,7,11,13,17,19,23,29}; map m = {{0,0}}; int maximum; cin >> maximum; if(!cin || (maximum < 1 || maximum > 99999)) { cout << "Wrong input. Using default value" << endl << endl; maximum = 100; } mt19937 rg; rg.seed(random_device()()); uniform_int_distribution dist(1,maximum); cout << "Range of numbers : (" << 1 << "," << maximum << ")" << endl << endl; for_each(factors.begin(), factors.end(), [&](int i) {m[i]=0;}); while(true) { int n = dist(rg); int max_factor = 0; for_each (factors.begin(), factors.end(), [&](int i){ if(n%i==0) max_factor = max(max_factor,m[i]++); }); if(!max_factor) max_factor = m[0]++; if(max_factor==screen_width) break; } for_each (factors.begin(), factors.end(), [&](int i) { cout << setw(2) << i << " " << string(m[i],'*') << endl;}); cout << " 0 " << string(m[0],'*') << endl; return 0; }

[challenge] graph factors