Yesterday i participated in this contest On UVA OJ from Bangladesh

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=13&page=show_contest&contest=253

I am here to discuss about

http://uva.onlinejudge.org/contests/253-dead8a0b/11806.html

Am i right? The problem can be solved considering the following cases

Case1 : There is non-zero solution only when

Case2 : The arrangment can be done in following way:

Thats'all

Please comment whether this can provide a solution or not

One thing more at the UVA site can't i submit the solution to this problem any more?

Thanks for ur response

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=13&page=show_contest&contest=253

I am here to discuss about

**problem C(Cheerleaders)**which i couldnot submit. The first two problems A and B were easier for me.__Problem C link__http://uva.onlinejudge.org/contests/253-dead8a0b/11806.html

Am i right? The problem can be solved considering the following cases

Case1 : There is non-zero solution only when

**K>=2**. otherwise zero solution**if Case1 is true, then**Case2 : The arrangment can be done in following way:

**I**. place 2 Cheerleaders on the one of the opposite corners and the rest leaders can be put in the rest of N*M-2 cells .**II**place 2 Cheerleaders on other opposite corners and the rest leaders can be put in the rest of N*M-2 cells**III**place 2 cheerleaders in adjacent corner and 1 leader on one side and the rest in the remaining N*M-3 cells**IV**III way can be repeated 3 more times**V**place 3 cheerleaders in 3 corners and the rest in N*M-3 cells**VI**V way can be repeated 2 more times**VII**place 1 cheerleader(it will cover 2 sides) on one of the corner and 2 leaders on the other 2 sides and the rest leaders on N*M-3 cells**VIII****VII way**can be repeated 3 more times as we have 4 corners in total**place 4 cheerleaders on all the 4 sides aand the remaining cheerleaders on the N*M-4 cellls**

IXIX

Thats'all

**should be appropriately dealt as sometimes (k has to be)**

KK

**I**K>=2**II**K>=3**III**K>=4Please comment whether this can provide a solution or not

One thing more at the UVA site can't i submit the solution to this problem any more?

Thanks for ur response

My solution:

First find out all the points on four sides. It will have (n + m) * 2 - 4 points at most.

Every point can cover some sides. We use a binary number to describe the 'ability' of every point.E.g. If a point can cover sides 1 & 3 , then we give it a number 2^0 + 2 ^ 2

Then use dynamic programming: F[S][tot][Mark] means how many ways are there if we put tot cheerleaders on first S points and the state of the sides is Mark.....

The things following is too simple :)