Could some one provide me the solution of the problem " World Record" for the April Long Challenge (2021) of Division 3.
The code created by me has some issues I don’t know what , check it out if it worth to watch
using namespace std;
float one(float *deci)
float value = (int)(*deci * 10);
return value / 10;
float two(float *mal)
float value = (int)(*mal * 100);
return value / 100;
j = vk1k2*k3;
t = 100/j;
dude u can check editorials for better understanding
Bro I didn’t found anything useful yet.
Bro your code too was submitted but could not satisfy the subtask.
Try to put this input set to your code (test case = 1 , k1 = 1.03 , k2 = 1.04 , k3 = 1.05, v = 10.21) it should print “No” but it will not .
The point is I don’t have any problem in submitting my code, its just the fact I wanna pass all the constraints given in the problem.
If you observe my code I’ve put nested if be the calculation of “t” just to pass all the constraints , I’m still wondering what went wrong.
Bro it should print “YES” because the multiplication if k1 * k2 * k3 * v = 11.4837996 and when we divide it by 100 :- 100/11.48 = 8.71 sec. This means he can beat the world record. So the answer should be YES
HImanshu bhai , I don’t know if I’m interpreting it right , but the problem constraints clearly says k1,k2,k3 contains one digit after decimal, that mean no matter if I put 1.0 or 1.03 or 1.054 , it should consider it as 1.0 only. (Restricting the decimal , PS that’s what I’ve done in my code before the main function)
Did your code passed the sub task too.
Try this input set (test case = 1 , k1 = 1.03 , k2 = 1.04 , k3 = 1.05, v = 10.21)
But its not supposed to … cause the constraints clearly says k1,k2,k3 contains one digit after decimal and v contains 2 after decimal. (we have to restrict the decimal)
Thus it doesn’t matter if I put 1.0 or 1.02 or 1.034 , it should consider it as 1.0 only.
this ig is the most straight forward way to solve it ^^, truncation and rounding off of cases are taken care of here