Simulate queue using two stacks.
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@swayambhoo:
ofcourse a cubical room must me symmetrical at allcorners, Hence, neway it
will reach in
min_dis=sqrt((4+3)^2+5^2)=8.6
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For n x m take first element of all the rows and insert in a min heap.
Now take the smallest element and and if we have a track of from which row
it belongs, we can take an element out of that row
and insert in the heap.
this will be done n x m times. Hence we have a time complexity of
O(nmlog(m))
a == b == a b - EPS
a = b == a b + EPS
all other relations are straighforward.
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char type is signed by default, so it represents value 255 as -1.
Code outputs first byte as unsigned int value, where -1 == 4294967295.
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i was doin a check on which one of the two floats is greater using sign
but was unable to do soi m actually solving this prob:
http://www.codechef.com/JAN11/problems/FLUSHOT/
http://www.codechef.com/JAN11/problems/FLUSHOT/i hav code wid me shud i
post it here?
On Sun, Jan 2, 2011 at 9:13
i dont know how shud i modify my code using epsilon shall i mail it to
someone or straightaway post it here .?
On Sun, Jan 2, 2011 at 9:58 PM, sanchit mittal sm14it...@gmail.com wrote:
i was doin a check on which one of the two floats is greater using sign
but was unable to do soi m
There is no need in epsilon comparison when you need to compare it with or
operators.
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#includestdio.h
#includeconio.h
int main()
{
int t,D,N,i;
float T,*x,*y,*z,maxNo,minNo;
scanf(%d,t);
while(t--)
{
scanf(%d %f,N,T);
x=(float*)malloc(sizeof(float)*(N+1));
y=(float*)malloc(sizeof(float)*(N+1));
oups gt my mistake.btw hnx for replying
On Sun, Jan 2, 2011 at 10:10 PM, juver++ avpostni...@gmail.com wrote:
There is no need in epsilon comparison when you need to compare it with
or operators.
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http://www.codechef.com/JAN11/problems/FLUSHOT/
*m getting wrong answer*
*if there is any problem wid my algorythm do let me know*
#includestdio.h
int main()
{
int t,D,N,j,i;
float T,*x,*y,*z,maxNo,minNo;
scanf(%d,t);
while(t--)
{
scanf(%d %f,N,T);
@Salil: Just to make sure we are on the same page, A hits with 100%
probability, B hits with 50% probability, and C hits with 33%
probability. C shoots first, then B, then A. Then the shooting
continues among the survivors in that order until only one is
standing.
If all three are alive and it is
Wow: Its a Running Contest :)
On Sun, Jan 2, 2011 at 11:07 PM, sanchit mittal sm14it...@gmail.com wrote:
http://www.codechef.com/JAN11/problems/FLUSHOT/
m getting wrong answer
if there is any problem wid my algorythm do let me know
#includestdio.h
int main()
{
int t,D,N,j,i;
This link may be helpful:
http://www.cygnus-software.com/papers/comparingfloats/Comparing%20floating%20point%20numbers.htm
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@Dave
1. In the end only 1 will survive (after max of 2 rounds).
i.e. P(A survives in end) + P(B survives in end) + P(C survives in end) = 1
2. Now, by shooting in air C increases his probability by your argument,
which is not good for A.
Thus, P(A shooting at C) should NOT be 0 if A is
Solve it by yourslef.
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@juver++ how will implwment find_min() function?
On Sun, Jan 2, 2011 at 2:33 PM, juver++ avpostni...@gmail.com wrote:
Simulate queue using two stacks.
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@Salil: If C shoots at A instead of into the air, he increases the
odds that he will be shot by B, because if C hits A then B will shoot
at C instead of A.
On the other hand, if C shoots at B instead of into the air, he
increases the odds that he will be shot by A.
Thus, shooting at either A or
keep min for stack is easy. just use another stack to keep the min for each
top.
Sent from Nexus one
On Jan 2, 2011 11:43 AM, Anuj Kumar anuj.bhambh...@gmail.com wrote:
@juver++ how will implwment find_min() function?
On Sun, Jan 2, 2011 at 2:33 PM, juver++ avpostni...@gmail.com wrote:
@Anand:I went through the link posted in your blog.But I found the
method little bit hard to understand.
@Aviral:Please elaborate the approach or give some link as in your
blog I didn't find the solution.
It will be very helpful.Thanks in advance.
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On Sun, Jan 2, 2011 at 8:30 PM, Akash Agrawal akash.agrawa...@gmail.comwrote:
I have written a kinda messed-up code for the same. Which is basically a
bottom-up approach.
Please find the same as attached. Some boundary conditions might be missed
and code can be written in a more decorated,
Hello Anand
I am getting the error Blog not found . Could you please provide me
the correct link.
-Prims
On Dec 31 2010, 1:44 pm, Anand anandut2...@gmail.com wrote:
Longest increasing subsequence using segment tree with O(nlogn)
@Dave
In point # 1, I had mentioned that P(A survives) + P(B survives) + P(C
survives) = 1 as the dual will be carried out till only 1 man is left. Do
you agree on this?
By your calculations, P(A survives) = 0.17 * 1 + 0.5 = 0.67
P(C survives) = 0.4962
They add to more than 1. Please let me know
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