OJ Board

**SPOILER**
there is a simple dp solution
the state is (n,last) where n is the number of hourses left and last is wether the previous house is connected or not
now each i have 3 choices to made
1.connect the current house with the center
2. connect the current house with the next house
3. if the ...

My Acc prog gives the following o/p for these cases

2 ...

if you really understand what sieve is doing then it's a very easy problem ...
just do sieve 2 times ... (more explanation would be spoiler )
i just reimplemented sieve with small changes and i got rank 5 in this problem after small optimization of my code ...

there is a very simple Algorithm to get the number of zeros without making any prime factoring ...
All what we have to do is finding how many times the base divides the factorial ... we can do it in O(n) but we will have to avoid the overflow ...
that's where the modulus part comes ... and i will ...

for people who's still getting WA even with the previous I/O is OK
try this case ...
ur output may be
1 1
but the right answer is
2

hope this helps

According to my Acc prog the Ans is
4
5
2
3
4
3
5
4
4
3
1
2
2
1
3
3
4
4
7
4
4
1
3
3
2
1
5
2
4
2
5
3
2
6
4
9
3
3
3
2
3
3
1
2
2
3
4
3
3
3
1
1
4
6
2
3
3
3
3
2
4
4
3
3
4
2
2
4
4
2
2
3
2
3
4
4
5
1
4
2
5
5
4
2
3
3
4
5 ...

i have a question guys ... how can i get this 0.000 time ?!

dennywithy wrote:why i get wrong output at this test:

from the problem statement
(The distance lost to fatigue is always 10% of the first day's climbing distance.)

keep this in mind

can someone tell me why i was getting TLE when my array was for the first 5000 and when i changed it to 5050 it worked in 0.6 ?!!?!?!!?!

Hi,
to solve this problem u need a clever way to generate the number of factors of factorial -( solve the problem 884)-
for each number in the range 1 -> 10 000 000 ... maybe the Algorithm used in the problem 884 isn't enough for this problem.... but it will help you to of course.
then it's easy to ...

for anyone keep getting WA
just take care of the 'S' in case the output is greater than or equal to 1
and try this case:
Input: Output: Ahmad

it would be great if zoranh (or anyone) helped us by the Mathematical principle behind this conditions to check .... this is why we solve such problems i guess .. to learn

someone tell me why am getting wrong answer or i will sucide
here's the java code
import java.util.Scanner;

public class Main{
public static void main(String[] args) {
int r = 1000000;
boolean[] visited = new boolean[r+1];
for (int i = 2; i < r; i++) {
if (!visited[i]) {
for (int j = 2; i ...

i think the problem is about the i/o ... so please help me cuz i'm not able to figure it out
i tested my code alot of times and i always find the expected answer ...
here is my implementation using java

i think the problem is about the i/o ... so please help me cuz i'm not able to figure it out
here is my implementation using java