problem with implementing this question of codechef

[DGLCfSEw]

@codexhammer, why you got continue in an if else statement? else will not run after if anyways.

[ffloyd]

@MonkeyD
show me your coefficients along with output here otherwise you can PM me, okay

[shadow98]

.

[DGLCfSEw]

@ffloyd you can see your PMs

[ffloyd]

@MonkeyD

ohkay i got ,BTW you have cool variable name lol ;)

[closed account 3UohqMoL]

Can anyone join this thread to help me out ?

http://www.cplusplus.com/forum/beginner/238075/

[DGLCfSEw]

@ffloyd please check your Inbox once again.

[yousuckloljk]

Codechef authority here..tadaa... :D
jokes apart..!!

meanwhile for all who are worried about plagiarism..
well, the MOSS technique works in many ways. its hard to say what sort of technique codechef uses for their system.

potential solution:
irrespective of origin coding language, implement same code in different language. (which ultimately generate different intermediate level code.) and thus we can be safe from plagiarism detection system.

telling it from my previous experiences on same platform :P

[spam291203]

@MonkeyD have you solved the problem successfully?

[codexhammer]

@shadow98 No I took the input normally. I didn't do anything different for the second subtask.

[keve]

hey guys anyone who has done the solution for Vision problem.

please share the code guys.

[43hf216C]

@keve if you can give me code for sheokand i can give this one to you

[closed account N6Xyb7Xj]

@texasgirl switch on your private message i will give it to you..!

[keve]

@texasgirl i will give it to you the code of sheokand
tell me how to give the code to you

[ffloyd]

@MonkeyD
double time1= -y/(2*x);

double time2= ( sqrt(y*y-4*x*z) )/(2*x);
limitation in PM
i think you can figure it out be conscious of plag

[keve]

@texasgirl
please enable your private messages
i will give you the code on that
is it OK.

[43hf216C]

yes pm me

[4iEqDjzh]

@ffloyd

I don't get it, you mean to change
T1=(-B+sqrt(B*B-4*A*C))/(2*A);
T2=(-B-sqrt(B*B-4*A*C))/(2*A);
to

double time1= -y/(2*x);

double time2= ( sqrt(y*y-4*x*z) )/(2*x);
????

[ffloyd]

@dMonkey
i just split the time in two parts
here A=x,B=y,C=z
now you can calculate
T1 and T2 according to you
by time1+time2 and time1-time2 respectively
cann't PM you , i cross the limit ;)

[keve]

@texasgirl
see your PM and i have send

now please give me the code for Vision.