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Posted: Tue Nov 02, 2004 12:30 am
I use a recursive formula:
f(n, k, m) = sum (1 .. m) f(n - i, k - 1, m)

Posted: Tue Nov 02, 2004 1:25 am
That's it??!?!?!

And here I am doing it in such a complex way. I got every single valid sequence and then found all the combinations/permutations (never knew the difference ) of them...

### 10721 - TLE

Posted: Mon May 23, 2005 2:46 pm
I have TLE for this problem, because I use a Formula to calculate recursively the final number for this problem. How can I quit TLE?

Posted: Mon May 23, 2005 3:49 pm
If you solve it by a recursive formula, instead of coding a simple recursive function, you should store the value you get into a table so that you don't need to compute everything again and again each time.

### I have TLE

Posted: Mon May 23, 2005 4:34 pm
My problem persist. I don't know how to reduce time.

This is my code for this problem

Code: Select all

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

void Formula(int tot,int sep,int max){
int i;

if (tot<sep) return;
else if ((tot<=max)&&(sep==1)) { contador++; return;}
else if (sep==1) return;

for(i=1;i<=max;i++) Formula(tot-i,sep-1,max);
}

int main (int car, char** v) {

char linea[100];
int total,separaciones,maximo;

while(fgets(linea,100,stdin)!=NULL){
sscanf(linea,"%d %d %d",&total,&separaciones,&maximo);

Formula(total,separaciones,maximo);

}

return 1;
}
``````

Posted: Mon May 23, 2005 5:09 pm
You misunderstand what I mean. You should store the value in a 3D-table like:

Code: Select all

``````int table[50][50][50];

int Formula(int tot,int sep,int max){
if(table[tot][sep][max]==-1)
{
// compute the value of Formula(int tot,int sep,int max)
// recursively here and assign it to table[tot][sep][max]
}
return table[tot][sep][max];
}

int main (int car, char** v) {
char linea[100];
int total,separaciones,maximo;

initialize all entries of the table to be -1;

while(fgets(linea,100,stdin)!=NULL) {
sscanf(linea,"%d %d %d",&total,&separaciones,&maximo);
printf("%ld\n", Formula(total,separaciones,maximo));
}

return 1;
}``````
This is a standard technique called dynamic programming. If you are computer science major, you will learn it in some intermediate/advanced algorithm course.

Posted: Mon May 23, 2005 8:15 pm
Hello.
This problem also can be solved using 2D matrix or just can be found a formula using PIE.
Eduard.

Posted: Sun Jun 19, 2005 11:40 am
Can you explain a bit what's the 2D approach. I've done it using a 4D array with dimensions [50][2][50][50].

Regards
Sanny

Posted: Sat Dec 15, 2007 5:48 pm
I have used 3D memo .. I am getting WA ..
can u give me some I/O where my code fails ..?? Thanks in advance ...

Code: Select all

``removed after AC..``

### Re:

Posted: Mon Oct 19, 2009 10:36 am
I matched the very all outputs from Krzysztof Duleba and I get WA...
Why would that be?

### Re: 10721 - Bar Codes

Posted: Sat Aug 02, 2014 5:42 pm
Hints:
-------
Complete search over all bar widths, in each recursive call you should try to add a bar with all allowed width.
Recursively do that until all your bars width = n and number of bars = k Here you found a solution.

You can use a 2D array to store answers. in order not to re-calculate any of them (DP).

### Re: 10721 - Bar Codes

Posted: Tue Mar 17, 2015 7:01 am
got ac on 1st submission actually 2D array is enough . as 'm' is not changing so we don't need 'm' as dp state