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| If "wheel can be rotated only counter-clockwise" | ძამაანთ [Tbilisi SU] | 1370. Волшебник | 11 авг 2013 03:47 | 1 |
then how the heck does it happen that "after one click the first digit will go out of site and the 11-th digit will become visible."
Edited by author 11.08.2013 03:47 |
| Why WA2? | Alex Vistyazh [Ivye School] | 1815. Ферма в Сан-Андреасе | 10 авг 2013 21:39 | 2 |
Why WA2? Alex Vistyazh [Ivye School] 7 авг 2013 17:02 I can't find my mistake... Please give me second test. Sorry, but i find my mistake and i have AC :) |
| Does brute force get AC? | Sergiu Repede | 1052. Охота на зайцев | 10 авг 2013 02:42 | 3 |
Yes you can take any tow points and count all other coliniar points.
Edited by author 25.09.2004 21:19 Yes. I have just got AC with my O(n^3) algo. But i used c++. |
| dunno why but I got AC XD | ძამაანთ [Tbilisi SU] | 1214. Странная процедура | 9 авг 2013 19:56 | 1 |
The procedure given in the problem is the solution :D
my code:
long x,y;
cin>>x>>y;
P(x,y);
cout<<x<<" "<<y;
XD |
| some test data added? | suiyuan | 1976. Оптимизация игры | 9 авг 2013 19:37 | 1 |
|
| Can I solve it considering every point? | ძამაანთ [Tbilisi SU] | 1046. Геометрические грёзы | 9 авг 2013 19:00 | 1 |
huh?
just yes or no please : ] |
| Question to Authors about several answers | Alex Danilyuk [SESC USU Dandelion] | 1730. Челмедведосвин | 9 авг 2013 14:08 | 1 |
input:
man-bear-pig
1/2 1/4 1/4
output:
7
man
man
pig
bear
1 2
3 4
5 6
input:
man-bear-pig
1/2 1/4 1/4
output:
5
man
bear
pig
2 3
1 4
Are both of answers correct?
input:
man
1/1
output:
7
man
man
man
man
1 2
3 4
5 6
Is this answer correct? |
| TL1, WTF? | PrankMaN | 1964. Китайский язык | 9 авг 2013 13:52 | 1 |
TL1 with G++ but AC 0.015 with VC++ 2010? My solution is only 14 strings and only one cycle to read ai. |
| Let open discussion | svr | 1682. Чокнутый профессор | 9 авг 2013 13:30 | 5 |
Could anybody correct answers given by my Wa program:
1 3
2 5
3 5
4 7
5 7
6 8
7 10
8 12
9 13
10 13
11 14
12 15
13 17
14 19
15 20
16 23
17 21
18 22
19 15
20 24
21 26
22 18
23 30
24 32
25 32
26 31
27 23
28 33
29 13
30 35
31 37
32 39
33 29
34 42
35 43
36 44
37 43
38 44
39 45
40 46
41 47
42 48
43 34
44 39
45 54
46 56
47 58
48 48
49 27
50 57 Here are the correct answers:
1 3
2 5
3 5
4 7
5 6
6 8
7 10
8 12
9 11
10 11
11 8
12 15
13 14
14 17
15 20
16 23
17 18
18 20
19 15
20 24
21 24
22 18
23 21
24 32
25 19
26 27
27 23
28 31
29 13
30 35
31 22
32 39
33 29
34 35
35 40
36 44
37 30
38 33
39 42
40 44
41 31
42 45
43 32
44 39
45 50
46 43
47 44
48 60
49 27
50 39 Thank! Debugging in range[1..50] was enough for AC. Edit: Nevermind
Edited by author 03.12.2009 01:15
Edited by author 03.12.2009 01:15 |
| I got ac.But it's clearly that I'm wrong. | neko13 | 1807. Патроны для Максима | 9 авг 2013 01:21 | 2 |
My ac program print only 500 when n=625.
I think the test number is not good enough. I'm sorry.I use the wrong project. |
| What's mean this problem? | stat958 | 1427. SMS | 8 авг 2013 08:23 | 1 |
This is very important: "SMS message which consists of latin letters and spaces only can be up to M characters long while the length of SMS message which consists of any characters is limited by N characters." Because non-latin letters are coded by another charset and message header will took 5 extra bytes. Read the problem task about N and M you will understad; test case: 3 5 aaaa 答案是 1 3 5 aaa! 答案是 2 http://blog.csdn.net/zhangyanxing666/article/details/9831623 |
| Give me a hint, I AC this using preprocessing | monyura[ONU 1 2/3] | 1807. Патроны для Максима | 7 авг 2013 22:25 | 8 |
I use Dynamics that originally need 31607*3400 operation for input 31607*31607, but when I notice that only first 114 prime numbers uses, and only first 8 of them with power more than 1, I reduce operation count to 200*31607. BTW it works smth about 3 sec. So I use preprocessing to solve it. Does exist another algorithm, or I need to optimize this one? (I think people, who ac this problem, guess algorithm by it complexity) Maybe that can help someone to solve...
/*
* Firstly note that according to the problem statement N is composite.
* Then find smallest divisor M that N%M == 0 and let G = M/N.
* Now we can factorize N as N = G*M = G*(a1 + a2 + ...) and consider
* values G*a[i], gcd(G*a1, G*a2, ...) == G as the answer. Those values
* will have the largest GCD. It can be easily proved that to maximize
* LCM all a[i] must be coprime and be of the form prime^k.
* Use dp to find such partition...
*/ But when N~ 10^9 and N - is prime number, than smalles divisor M = N, and G = N/M = 1.
a[1] + a[2] + ...a[k] = M ~10^9 . how do use DP? N cannot be a prime number, it's always composite. Read the statement carefully
> each box contained at least one hundred cartridges.
> Anka noticed that there was the same number of cartridges in all boxes
maybe it will be more correct to replace "in all boxes" with "in each box"?
Edited by author 11.02.2012 17:56
Edited by author 28.02.2012 20:32 N cannot be a prime number, it's always composite. Read the statement carefully
> each box contained at least one hundred cartridges.
> Anka noticed that there was the same number of cartridges in all boxes
maybe it will be more correct to replace "in all boxes" with "in each box"?
Edited by author 11.02.2012 17:56 Thank you!!!!!!!!!! Hovewer, if M - is smalles divisor of N, than 1 < M < 31623 ! I Ac, with 0.046 s !! Edited by author 28.02.2012 22:19 Edited by author 28.02.2012 22:19Could you give me some hints about how to breakup a prime as m=a1+a2+a3... so that lcm(a1,a2,a3,...) is biggest? |
| 197*197 | Alexandr172 | 1807. Патроны для Максима | 7 авг 2013 22:21 | 4 |
197*197 Alexandr172 3 май 2012 21:27 Please, write solve for N = 197*197...
My result: 1576 1773 4925 1379 2167 2561 3349 3743 4531 5713 6107 197 197 197 197 197 Can you give me some hints about how to breakup a prime as m=a1+a2+a3... so that lcm(a1,a2,a3,...) is biggest? Can you give me some hints about how to breakup a prime as m=a1+a2+a3... so that lcm(a1,a2,a3,...) is biggest? |
| note for WA#1 | hoan | 1283. Гномик | 7 авг 2013 18:36 | 3 |
use ( > ), and don't use ( >= ).
GOOD LUCK |
| I need help | Davrbek Tursunmurodov | 1001. Обратный корень | 7 авг 2013 18:05 | 1 |
I have recently started programming,please help me. where is wrong in this program?
It was compilition error.
#include<iostream>
#include<math>
double b[500000];
int main(int argc, char* argv[])
{ unsigned long long a,i=0,j;
while(a!=EOF) {std::cin>>a;
i++; b[i]=sqrt(a); }
for(j=i;j>1;j--)
std::cout<<b[j];
return 0;
} |
| please help me? | Adhambek | 1001. Обратный корень | 7 авг 2013 17:50 | 3 |
how many do values enter in this program? I don't know . How do i work?
you can use global array to solve it,but it should be large,at least 500000.
for example : double answers[500000]. Did you find answer? Please,write here Adhambek.
Edited by author 07.08.2013 17:53 |
| WA # 7 | R. Dubinin | 1878. Кубик Рубинчика | 7 авг 2013 17:30 | 1 |
WA # 7 R. Dubinin 7 авг 2013 17:30 please give me some tests
Edited by author 07.08.2013 17:31 |
| WA4 and WA8 | Plamen_N | 1929. Не все любят плюшевых мишек | 7 авг 2013 07:54 | 2 |
Test#8 -> N/3 == M and Holden is not hater => output is M*(N-M-1).
Test#4 -> N/3 > M => output is '0'. what is the formula when n/3<m? |
| Have wa #12 | cap_prot | 1699. Черепахи и повороты | 6 авг 2013 19:17 | 2 |
My program is more advanced than LCA. It uses BFS and could work on arbitrary graphs. Could somebody help with this test? I know that if there is no bug with my program then there is a problem with connectivity of graph from start point to end point. Check for cases when start = end.
also don't use incorrect comparison. for example:
newPoint = point+delta
if(0<=newPoint && newPoint<w*h) |
| Delete this post ( double post ) | begi | 1028. Звёзды | 6 авг 2013 17:08 | 1 |
Double post, please remove this one.
Edited by author 06.08.2013 17:09
Edited by author 06.08.2013 17:14 |
