atique Query has double type input instead of integer. This is not matching problem description. That's bad. That is why statistics of the problem shows less acceptance. Please fix.
GustavBertram Actually, it just doesn't fit inside an int. It's all there in the problem statement. q <= L^2 and L<=10^9, so q<=10^18. LONG_MAX in this environment is 9223372036854775807, so that is juuuust sufficient.
masotrix You are the one right man.
atique I am sorry. I made wrong guess. You are right.
sgrebenkin Yeah. Use long.
StoneCold_Xi [deleted]
protojas For python users having trouble with JUST number 4 - try doing
format(answer, '.20f')to get the right precision required.
dbhrockzz import math l,s1,s2 = map(int,raw_input().split()) n = input() q = [] for i in range(n): q.append(input()) rel = abs(s1*math.sin(math.radians(45))-s2*math.sin(math.radians(45))) for qi in q: if qi==(l**2): print 0.0000000 else: print format((l-math.sqrt(qi))/rel,'.20f')
Test Case 4 is still coming wrong
jane_beguin with this code you have O(1) complexity and it works
import math t, v1, v2 = map(int, input().split()) for _ in range(int(input())): a = int(input()) rep = math.sqrt(2)*(t-math.sqrt(a))/abs(v2-v1) print (rep)
sgrebenkin Oh c'mon, sin(Pi/4) = sqrt(2)/2;
bigbang1498 Can you please explain the logic behind using format here, I wasn't getting 4th testcase right but after doing what you suggested I got it right
protojas I noticed that for people using C++ they were having precision errors.
format(answer, .20f)just makes the answer go to 20 decimal points, which is somehow required for question 4.. not sure why thoughbigbang1498 thanks
RahulHP Thanks a lot, I wasted a submission due to this before reading your comment!
tvanloon For those using C++ stuck on test cases 3 and 4, use cout << std::fixed <<
withasingle_t Thanks!
q1a2z3 What does that do?
adarsh97 you are awesome bro....very very thank you.
metamemelord cout<<setprecision(50)<<ans<<endl;
works as well!
manish_chauhan i m not getting it can someone explain me logic of this progam
JasolAsked to answer Hm, doesn't look like I can paste a picture in here, but I'll try to give you some information that helped me. You're trying to solve for time, so you want an equation with time where you can isolate time, and solve it. Distance = speed x time. There are two distances available to you, the distances of the bottom left corner of both squares. Using this information, and the picture provided with the problem, you can solve for time.
bismitapadhy I am not able to understand the question exactly.can you please explain
subash_kotha2 Working C++ Code
#include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> using namespace std; int main() { double l,s1,s2; cin>>l>>s1>>s2; int q;cin>>q; do{ double area; cin>>area; if(area<=l*l) { cout<<(sqrt(2)*((double)l-sqrt(area)))/abs(s1-s2)<<endl; } else cout<<0<<endl; q--; }while(q!=0); }
shahalamansari98 could you please explain the code?
pl_lepko I solved this problem by solving: Quadratic equation...
q= ( L+(time*p)) * (L+(time*p)); (L+t*p) * (L+t*p) = q q = L^2 + 2*L*t*p + t^2*p^2; q - L^2 = 2*L*t*p + t^2*p^2; q - L^2 = 2*L*t*p + t^2*pS; (a)*t^2 + (b)*t - (c) = 0; //quadratic equation....
masterpradeep Getting runtime error for testcases 4,5,6. suceesfully executed 1,2,3.How to find out the reason for run time error. Help
public class Solution {
public static void main(String[] args) { Scanner g=new Scanner(System.in); int side=g.nextInt(); int v1=g.nextInt(); int v2=g.nextInt(); double a = Math.sqrt(2); double diagonal=a * side; int v3=Math.abs(v1-v2); int t=g.nextInt(); for(int i=0;i<t;i++) { int distance=g.nextInt(); double diagonal2=a * Math.sqrt(distance); if(v3 != 0) { double result = (Math.abs(diagonal-diagonal2))/v3; System.out.println(result); } else { System.out.println(0.00000000000000); } } }}
masterpradeep Solved it by using - nextDouble()
romario_yakunin This problem is sopposed to be solved with a formula.
cgira try changing your else statement to return infinity instead of 0.0. The logic is that if the two squares move at a same pace, they will never leave q area as shown on the picture
pGxplorer no need to write else part. It is already said that S1 and S2 will never be equal.
vvbhandare Will you please explain how you calculate this to get q? I tried to understand your code but could not get it.
ThanujaM Use long for v1,v2 and v3
ebrucecfa U should use
double distance=in.nextDouble();
int will result in Test Case failures in Java8.
bhuvanms why are we doing s1-s2 relative velocity concept??
rajatsaini Relative velocity will give us the speed of (original_diagonal - diagonal_of_qi).
poonia0312 hello your code give runtime error in testcase 3,4,5 plz correct it
a01071846617 numer 3,4,5 wrong answer
vector<double> result; for(int i=0;i<queries.size();i++) { if(s1>s2) { result.push_back(((sqrt((double)queries[i])-l)*sqrt((double)2))/(s2-s1)); } else { result.push_back(((sqrt((double)queries[i])-l)*sqrt((double)2))/(s1-s2)); } } return result; why........ teach me please....
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