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Discussion of Problem 1475. Ryaba HenBirne Ageev [USU] (Psych up club) What about light velocity? :D [1] // Problem 1475. Ryaba Hen 17 Nov 2012 01:45 Can be velocity lager than c? Ade Re: What about light velocity? :D // Problem 1475. Ryaba Hen 19 Aug 2023 10:42 Good Point! LoL yyll Newton // Problem 1475. Ryaba Hen 17 Jun 2021 14:46 Sir Isaac Newton is important to this problem. In TWO different ways. xurshid_n I long time try, and found this formula: [5] // Problem 1475. Ryaba Hen 17 Aug 2012 19:57 4*h * n ( n + 1 ) > H minimal n - is answer! but I doubt, that this is correct. It's correct or not? Shen Yang Re: I long time try, and found this formula: [4] // Problem 1475. Ryaba Hen 12 Oct 2017 06:23 v0*t*sin(theta)+0.5*g*sin(theta)*t^2 *n>=sqrt(H*H+l*l) v0=sqrt(2*g*h) t=2*v0/g tan(theta)=H/l. seems to be correct PS. this problem is too simple.. Edited by author 12.10.2017 06:25 Edited by author 12.10.2017 06:40 Shen Yang Re: I long time try, and found this formula: [3] // Problem 1475. Ryaba Hen 12 Oct 2017 07:50 the correct formula is 4*h*n*(n+1)>(H*H+l*l)/H Mahilewets Nikita [BSUIR] Re: I long time try, and found this formula: [1] // Problem 1475. Ryaba Hen 13 Oct 2017 00:29 Maybe provide some explanation Mahilewets Nikita [BSUIR] Re: I long time try, and found this formula: // Problem 1475. Ryaba Hen 13 Oct 2017 00:30 Oh Your previous comment is the explanation Felix_Mate Re: I long time try, and found this formula: // Problem 1475. Ryaba Hen 5 Aug 2018 12:45 Let O=(l,H), A=(0,0), B=(l,0). System of coordinates: OX=AO, OY = BO rotate on teta (teta=arctan(H/l)). Then: vx(t) = (v+gt)*sin(teta) vy(t) = (v-gt)*cos(teta) x(t) = V*sin(teta)+g*sin(teta)*(t^2)/2 y(t) = V*cos(teta)-g*cos(teta)*(t^2)/2 First point: (l, H) => x=0, y=0 Second point: y(t)=0 <=> t=2*v/g x(2*v/g) = 4*(V^2)*sin(teta)/g = d = 8 * h * sin(teta) (mg(H+h)=mgH+m*(V^2)/2) x = d, y = 0 Distance between First point and Second point = d Distance between Second point and Third point = 2*d ... Distance between i point and i+1 point = i*d If n = answer => d+2d+3d+...+(n-1)d<=sqrt(H^2 + l^2) d+2d+3d+...+(n-1)d+nd>sqrt(H^2 + l^2) Edited by author 05.08.2018 12:52 Abhishek Ghosh WA 2 // Problem 1475. Ryaba Hen 25 Aug 2014 01:35 I have derived an expression which is giving correct results for the input given on the problem page as well as the ones below : 2 1000000 1 1 1000000000 1 1 500 15811 I got this set from another thread on this problem, however I am still getting WA on test 2. I am rounding off the result in Python : round(abs(n)) and checking for the case when this will round to 0, which will happen for this case : 1 2 1 In which case I output 1. What's special about test 2 ? Can I get a similar test or an idea about the special conditions which have to be checked ? svr 1475 enforce me to study java and BigInt [2] // Problem 1475. Ryaba Hen 16 Aug 2007 15:55 1475 may be solved only in Integers with using big nums. But firstly we must be good pupils in physics and remember bright facts about falling spheres. Interestig that in floats this facts have so error disturbance that practically unseen. Igor Dex Re: 1475 enforce me to study java and BigInt [1] // Problem 1475. Ryaba Hen 25 Aug 2007 02:16 Please give us answers for input data 1000000 1 1 1000000000 1 1 svr Re: 1475 enforce me to study java and BigInt // Problem 1475. Ryaba Hen 25 Aug 2007 09:41 Program output 2 1000000 1 1 1000000000 1 1 500 15811
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