| ENG RUS | Timus Online Judge |
Обсуждение задачи 1335. Белый тезисMurodjon (TUIT Urgench) (n ^ 2 + 1)(2 * n * (n + 1) + 1) // Задача 1335. Белый тезис 31 авг 2016 01:58 A = n * n; B = n * n + n + 1; C = n * n + 1; A ^ 2 + B ^ 2 n^4 + (n ^ 2 + n + 1) * (n ^ 2 + n + 1) n^4 + n ^ 4 + 2 * n^3 + 3*n^2 + 2*n + 1 2*n^4 + 2*n^3 + 2*n^2 + 2 * n + n^2 + 1 2 * n*(n ^ 3 + n ^ 2 + n + 1) + n^2 + 1 2 * n ( n(n ^ 2 + 1) + (n ^ 2 + 1)) + n^2 + 1 2 * n((n^2 + 1)(n + 1)) + (n ^ 2 + 1) (n ^ 2 + 1)(2 * n * (n + 1) + 1) Evgeniy_Chernobrovkin(MUCTR-2013) Be attentive to take AC // Задача 1335. Белый тезис 21 янв 2014 03:33 It takes 15 minutes to me to understand, that output line have to be in decrease order! manishmmulani (n*n) (n*n + n + 1) (n*n + 1) =====> O(1) soln [6] // Задача 1335. Белый тезис 6 янв 2008 02:34 Rustambek_UWED Re: (n*n) (n*n + n + 1) (n*n + 1) =====> O(1) soln [2] // Задача 1335. Белый тезис 7 май 2008 17:44 Thank you very much for the formula. I can prove it. But had not found it. Desperados[KhAI] >> Starov_Lesha Re: (n*n) (n*n + n + 1) (n*n + 1) =====> O(1) soln [1] // Задача 1335. Белый тезис 15 июн 2008 15:58 Something very dark... The solution is more easy and more easy to understand too! Denis Koshman Re: (n*n) (n*n + n + 1) (n*n + 1) =====> O(1) soln // Задача 1335. Белый тезис 24 июл 2008 16:09 n^2+n, n^2+2n, n^2 Megatron Re: (n*n) (n*n + n + 1) (n*n + 1) =====> O(1) soln [2] // Задача 1335. Белый тезис 10 мар 2009 10:49 I can prove it . but i don't know how did you come out with that? IgorKoval(from Pskov) Re: (n*n) (n*n + n + 1) (n*n + 1) =====> O(1) soln [1] // Задача 1335. Белый тезис 3 дек 2011 03:37 Megatron писал(a) 10 марта 2009 10:49 I can prove it . but i don't know how did you come out with that? It's simple. Just c=n^2 ( by intuition ). So, find a and b. a=n^2+c and b=n^2+d ( because n^2 <= a,b,c <= (n+1)^2 = n^2 + 2*n + 1 ), where c and d is positive and c!=d( because a!=b(by text of this problem ) ) a^2 + b^2 = (n^2+c)^2 + (n^2+d)^2 = 2*n^4 + 2*n^2(c+d) + c^2 + d^2 Look at this 2*n^4 + 2*n^2(c+d) + c^2 + d^2. Each summand must be divisible on c=n^2. So, just, c = n and d = 2*n Answer: a = n^2 + n b = n^2 + 2*n c = n^2 P.S.: You can say just a = n^2 and analogous prove another formula. Edited by author 03.12.2011 03:41 LyanA Re: (n*n) (n*n + n + 1) (n*n + 1) =====> O(1) soln // Задача 1335. Белый тезис 21 янв 2013 20:11 i was very surprised it's just 3 strings of code you're perfect =) Farhad Ghasemi Pretty easy question // Задача 1335. Белый тезис 31 янв 2008 12:03 Don't brute force thats the trick ! :D PPaavveell Why WA test #2??? // Задача 1335. Белый тезис 28 апр 2007 20:57 PPaavveell Kak reshit' etu zadachu legkim sposobom??? // Задача 1335. Белый тезис 28 апр 2007 20:57 wangchun Is there an easy way to do it? [2] // Задача 1335. Белый тезис 2 ноя 2004 13:03 Edited by author 02.11.2004 13:03 Azazel Re: Is there an easy way to do it? [1] // Задача 1335. Белый тезис 23 мар 2007 18:27 Em..We can write n*n n*n n*n..2*n*n mod n*n=0..So why WA?! the.zett Re: Is there an easy way to do it? // Задача 1335. Белый тезис 30 мар 2007 21:30 output should contain three different integers
|