OJ Board

#include <stdio.h>

double det (double a11, double a12, double a21, double a22)
{
return a11 * a22 - a12 * a21;
}

double main()
{
int N;
double d,area,x1,x2,y1,y2,x3,y3,x4,y4;
double x,y;

scanf ( "%d", &N );

printf ( "INTERSECTING LINES OUTPUT\n" );

while ( N-- )
{
scanf ( "%lf %lf ...

I dont really know why u asked me for that? Or is it me u meant by "Mr.Mahbub" ?

I am sorry if u didnt meant me by that by name. But if u indeed ask me tha question my answer is tha same as the prevous reply made by somebody named 'suman' in this topic.

Thanks
Light

Well Sending Table will certainly have faster run but i am really astonished why a time limit question arises:

As i resend my code today i get acc in .00.555 seconds. I think this is not bad.


#include<stdio.h>
unsigned long i,j,temp,n,len,s,ans,max,k,x;

int main(){
while(scanf("%lu%lu",&i,&j ...

There is no doubt that t4 = s*4/sqrt(3)...Becoz it is written by the

dicoverer of the problem!
The result for t2 is s* (4*root(6) - 6*root(2)) which is also as stated by

David Candrell...

So the prob is for t3 and t6..where he stated that t3 = 2.108+ and t6 =

3.168+..which do not hav analytic ...

There is no doubt that t4 = s*4/sqrt(3)...Becoz it is written by the

dicoverer of the problem!
The result for t2 is s* (4*root(6) - 6*root(2)) which is also as stated by

David Candrell...

So the prob is for t3 and t6..where he stated that t3 = 2.108+ and t6 =

3.168+..which do not hav analytic ...

Hi,

Did u notice that this is a simple example of "Matrix Chain Mutiplcation" dp

algorithm? In case u have used that and still getting TLE...u may find this

code helpful :)


#include<stdio.h>
#include<values.h>

#define MAX 60
#define inf MAXLONG

int main()
{
long L,N,i,j,n,x,y,q,l,k,t,sum ...

#include <stdio.h>
#include <stdlib.h>

#define MAX 1000000

char a[27] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
int f[27];
char s[MAX];

int cmpf(const void *a, const void *b)
{
char *x = (char *)a;
char *y = (char *)b;

if(f[*x-65]!=f[*y-65])
return f[*y-65] - f[*x-65];

return *x - *y;
}

int main ...

Wats the idea?

I got WA in greedy approach :
Always taking the line with highest end_x which has a lower left_index than curent covered length..

Thanks
Light

As this problem has really no trick i think its not bad idea to post my code her...
u can compare ur output with mines..(by fc or else.)



#include <stdio.h>
#include <string.h>
#include <stdlib.h>

struct S
{
char cand[100];
char party[100];
int vote;
} list[20000];

int cmpf(const void *a ...

Input can be like
3 2
100 1

That is first > last...in that case u need to swap while finding the max value.
But be careful to print 'first' and 'last' (as per ur code) according to input serial (not swapped..).

Thats all
Light

It may sound srtange to many but this problem has a faster(!) solution than DP , in BFS!!

Here is my code :>> U can perform I/O check with it..



#define MAX 301
#define QMAX 20000
#define EMAX 41

struct Point
{
int x;
int y;
};


int main()
{
int c,case_no,i,s,m,head,tail,found;
Point e ...

Ya it can be shorten.
As i use c++ i am putting a pseudocode here:

Let the interval is i......j

max <-- 0

for i = i to j
do s <-- i
len <-- 1
while s not equal to 1
do if ( s is odd )
s <-- s / 2 + 1
len <-- len + 2 /*-Tricky PArt -*/
else
s <-- s / 2
len <-- len + 1
if ( i = 1 )
len ...

Well..this is a liiltle odd for me to describe things without a figure..

If u had correctly assumed about the red and green line ..>

the side of trinagle = s;
s = 4 * (2r) * cos (A) + 2 * d

d = distance from the touch point of { bottom right circle with base of

triangle} to the right corner of ...

Can any one tell me exactly wats the ouput format??
I mean how many gaps between the words ?
Or wats the value for a, b in print (%a.blf)...

I think it will be easier if someone can give me printing part of his/her code..

Thanks
Light.

Thanks to Adrian. But i was actually getting WA for a silly stupid mistake.
Thanks. I got acc now

Light.