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Algorithm
an inversion if A[i]
A[j] and i j
[1] http://www.geeksforgeeks.org/archives/3968
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are uncountably
infinite.
On Mon, Dec 5, 2011 at 4:26 PM, Aamir Khan ak4u2...@gmail.com wrote:
On Mon, Dec 5, 2011 at 3:31 PM, saurabh singh saurab...@gmail.comwrote:
I was wondering can we design a machine(Even hypothetical) that can
find a *perfect square root *of any integer thats given
it.
On 03-Dec-2011 11:30 PM, Dipit Grover dipitgro...@gmail.com wrote:
^ we need to count each permutation and its reverse together as one
possibility since both would result in identical mst.
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Algorithm
points
for a given point, in time better than nlgn.
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answer..
How to deal with this situation ?
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On Sun, Oct 30, 2011 at 1:17 AM, Aamir Khan ak4u2...@gmail.com wrote:
In a university, students can enroll in different courses. A student may
enroll for more than one course. Both students and courses can be
identified by IDs given to them. Design a data structure to store
students
student.
d) Return all students in a list for a given course.
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As far as modular arithmetic is concerned,
(x * z *z )%MOD = (x *(z *z)%MOD)%MOD = ((x%MOD)*(z%MOD)*(z%MOD))%MOD
But when you try to calculate *(x * z * z)%MOD*, the intermediate result of* (x
* z * z) *overflows the integer limit and hence gives the wrong result.
similarly, intermediate value
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Lets say n,m and p are three long long integers.
i want to calculate (n*m)%p.
If (n*m) overflows the limit of long long then the answer would be wrong.
Moreover if i do ((n%p)*(m%p))%p then also there is no certainty of getting
correct answer.
How should i do this in C++ ?
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0;
}
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@shiv : Correct Answer should be : 5C3 X 5 X 4 X3 = 600
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#includecstdio
#define MOD (int)1e9
using namespace std;
int main() {
int A = 33554432;
printf(%d\n,A*A);
printf(%d\n,((A*A)%MOD));
return 0;
}
How can i calculate, lets say last 9 digits of square of 33554432 ?
Thanks,
Aamir
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*Output:*
0
0
On Thu, Apr 14, 2011 at 7:01 PM, AAMIR KHAN ak4u2...@gmail.com wrote:
#includecstdio
#define MOD (int)1e9
using namespace std;
int main() {
int A = 33554432;
printf(%d\n,A*A);
printf(%d\n,((A*A)%MOD));
return 0;
}
How can i calculate, lets say last 9 digits
On Thu, Apr 14, 2011 at 7:40 PM, Akash Mukherjee akash...@gmail.com wrote:
search nd read chinese remainder theorem
please elaborate how can i use chinese remainder theorem.
On Thu, Apr 14, 2011 at 7:04 PM, AAMIR KHAN ak4u2...@gmail.com wrote:
*Output:*
0
0
On Thu, Apr 14, 2011 at 7
there is no need for long long i
think.
Dave
On Apr 14, 8:31 am, AAMIR KHAN ak4u2...@gmail.com wrote:
#includecstdio
#define MOD (int)1e9
using namespace std;
int main() {
int A = 33554432;
printf(%d\n,A*A);
printf(%d\n,((A*A)%MOD));
return 0;
}
How can i calculate
Try this...
#include iostream
#include cmath
using namespace std;
#define DIGITS 10001
void mult(int N,int pro[],int len) {
int carry = 0;
for(int i=0;ilen;i++) {
int temp = pro[i]*N + carry;
pro[i] = temp%10;
carry = temp/10;
}
if(carry0) {
pro[len] =
are going to calculate 8 power 33
u compute 8 power 16 and multiply with the same to get 8 power 32
then multiply with 8 to get the result
On Thu, Mar 24, 2011 at 1:04 PM, AAMIR KHAN ak4u2...@gmail.com wrote:
Try this...
#include iostream
#include cmath
using namespace std;
#define DIGITS
You could try http://domjudge.sourceforge.net/ and do customizations in its
source code only..
On Thu, Mar 24, 2011 at 1:22 PM, .bashrc saurab...@gmail.com wrote:
Hi
I have been working on a progrmming judge that can be deployed in my
college.I have already developed a prototype in bash.I
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