+ 1

Security(Hard) - lacks adequate test sample [code coach] here is why!

This code shouldn't pass https://code.sololearn.com/cZLH6rQ4mG4H/?ref=app This code doesn't even check for multiple guards, it passes because the test sample doesn't check for one situation which is: xTxxGxGxx$xxGxx Again the code saves one guard location, in case of multiple guards it just overrides the previous guard location so the code reads: xTxxxxxxx$xxGxx Which doesn't output the expected result, meaning this code fails.

28th Dec 2019, 1:16 AM
Hisham Mohammed
Hisham Mohammed - avatar
2 odpowiedzi
0
Interesting find! Yeah, there are only very few tests, and some important variations may be missing... You should also email it to them, it's more likely they'll see it (although I doubt anything will happen).
28th Dec 2019, 8:52 PM
HonFu
HonFu - avatar
0
Here is my code : //Created by Gaurav Kaushik #include <iostream> #include <cstring>//for strlen(str.c_str()) using namespace std; int itr = -1;//function retruns this variable int it(string str , char ch , int start){ ::itr = -1;//reset the value of it global variable int *n = new int; *n = strlen(str.c_str()); while(start >= 0 && start < *n) { if(str[start] == ch) { itr = start; break; } start++; } delete(n); return ::itr; } int main() { string input = "xxxGxTxx$xxGx"; input.clear();//clearing input before input getline(cin, *input); //continuous input to avoid blanklines int *n = new int; *n = strlen(str.c_str()); while( *n <= 0) { getline(cin,input); } delete(n); int flag =1,T = 0, S = it(input, '
#x27;,0);//there is only one $ so if(S != -1) { while(T >= 0 ){ T = it(input, 'T',T); if(T==-1) break; if(T < S){ for(int i = S; i > T; i--){ if(input[i] == 'G'){ flag = 0; break; } } } else if(T > S){ for(int i = S; i < T; i++){ if(input[i] == 'G') { flag = 0; break; } } } } } if(flag == 1){ cout << "ALARM" << endl; } else{ cout << "quiet" << endl; } return 0; }
29th Dec 2022, 10:30 AM
Gaurav Kaushik