When we are finding distance b/w P and Q (x2x1)^2 + (y2y1)^2 whole underoot.(for 2d case). Now according to constraint of coordinates (this value (x2x1)^2 + (y2y1)^2) can exceed max value possible to store)? asked 12 Jun, 00:00

As specified in the constraints that the absolute value of the coordinates can be at max 2 10^9 . Now if we take x2=2 10^9, x1=2 10^9, y2=2 10^9, y1=2 *10^9, so it's pretty clear the value
answered 12 Jun, 15:50
Use long double.
(12 Jun, 15:52)
@vijju123 What about JAVA , it can support at max 64 bits
(12 Jun, 15:55)
Will double not suffice? Did you try that?
(12 Jun, 18:47)
@vijju123 no need to use long double, even for C++ double would suffice.
(12 Jun, 18:53)
yeah @vijju123 and @souradeep1999 you are right
(12 Jun, 19:04)
Thats true, but long double afaik is more precise than double. I never trust floating points so thought better safe than sorry xD
(12 Jun, 19:46)
How can long double store more than 64 bits i.e >10^18
(13 Jun, 10:12)
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As that is 64 bit then i think it can also store same as c++ upto 2^641,thats nearly 18*10^18. answered 12 Jun, 18:32

long double or even double can be used. They have a huge range (around $1.7e^{308}$) (check here) even though they are 64 bits since they use floating point representation, i.e. have some bits reserved for exponent and some bits for mantissa. answered 13 Jun, 16:45

Its guaranteed that coordinates of Q never exceed $2*{10}^{9}$ during entire time.