Why does this get tle?
Is there any way to find out which test case a solution failed on? This was a pretty straight forward implementation but it got WA. http://www.codechef.com/viewsolution/4081801
Please help.Tried every possible case mentioned here.Still getting WA.
All the sample test cases are passing. Also the corner case of “.”. However it is still giving wrong answer. Could somebody help?
http://www.codechef.com/viewsolution/4108137
this is my solution. can anybody tell me what’s wrong with this solution. it passed all tricky test cases but still showing wrong answer.
http://www.codechef.com/viewsolution/4119925
My code works fine for all the input cases, I also tried my code with the tricky cases mentioned in the editorial but i am still getting WA. Can anyone please explain to me what the problem is.
[1]
[1]: http://www.codechef.com/viewsolution/4120495
I’m getting a correct answer for every corner case, for every cases I read in the comments, yet getting a WA.
[1]
Can someone help me to tell where is it that I am going wrong?
[1]: http://www.codechef.com/viewsolution/4123826
getting WA after trying several test cases… CodeChef: Practical coding for everyone …can someone help…!!
I am getting all test cases correct on my system still it is giving wrong answer …please suggest what i am doing wrong…
CodeChef: Practical coding for everyone
I’m just doing these for practice, but its really hard to call “.” a “positive decimal number”. Arguably its also hard to call “0” and “0.0” as a “positive decimal number” (“non-negative decimal number” would have been more appropriate), but I don’t think that “.” passes as a “positive decimal number” in any culture I’m aware of.
plz can anyone point out the error… 5l8Jy2 - Online C Compiler & Debugging Tool - Ideone.com
i spent lot of time over it…(>_<)
@m1sterzer0 after reading your answer above…out of curiosity i checked your submission for the problem… i dont know why its giving the wrong output instead of being accepted as right answer z1YyRR - Online C++ Compiler & Debugging Tool - Ideone.com
http://www.codechef.com/viewsolution/3999376
This is my solution it is passing every case mentioned in the problem and every corner case as well. What is the problem here?
I am getting wrong answer error… please help…
#include < iostream>
#include < string>
using namespace std;
int main()
{
int n;
cin>>n;
string enPass,Pass;
while(n>0)
{
int r,counter=0;
cin>>r; // no. of rules.
char c[r],p[r];
for(int i=0;i<r;i++)
{
cin>>c[i]>>p[i]; // taking each rule as input
}
cin>>enPass; // taking encrypted password as input
Pass = enPass;
for(int j=0;j<enPass.size();j++)
{
for(int s=0;s<r;s++)
{
if(c[s] == enPass[j])
{
Pass[j] = p[s];
}
}
}
int str_Size = Pass.size();
for(int q = str_Size-1; q >= 0; q--)
{
if(Pass[q] == '0') Pass.erase(q);
else if(Pass[q] == '.')
{
Pass.erase(q);
break;
}
else break;
}
for(int t=0;t<str_Size;t++)
{
if(Pass[t] == '0') counter++;
else break;
}
Pass.erase(0,counter);
cout<<Pass<<"\n";
n--;
}
return 0;
}
I cracked this after trying different print techniques for the output.
First, I print the op as separate characters(i.e every single S[i] using loop). This didn’t work out. Secondly, I print the op as integers(using S[i]-‘0’ except for ‘.’). This didn’t work out as well. Finally after going through the solution of @pummy02(CodeChef: Practical coding for everyone) I got a hint. I assigned the 'last+1’th char as null and printed the op as a string. This worked out!
Can somebody help me with this solution, this gives correct answer for all the given tc but on submission gives wa.
https://www.codechef.com/viewsolution/14346300
@ar7thecoder you could have wrote it better. you are calculating leading and trailing zeroes for every string element which will give you a TLE. you can first decrypt the string and then calculate leading and trailing zeroes for the whole decrypted string. hope you get it now
1 1 k l books not working for this test case, your code
did the test cases include many mappings to ‘.’ ??
Shortest representation of 000.000 should be 0. Empty string can not be a valid representation.