@rgap
i think u have misunderstood the question
u are solving for old version of game
it says solve for new version specified in 2nd paragraph
but not a big problem, can be done with slight modifications
as input is small, just generate all B^2 possibilities and keep track of
absolute differences generated
if all differences are generated return true
else return false
On Tue, Jan 25, 2011 at 1:47 AM, rgap <rgap...@gmail.com> wrote:
> Here is my implementation ;)
>
> #include <iostream>
> #include <cstring>
> using namespace std;
>
> int balls[91]; //maximum balls
> int N;
>
> //verify if its impossible to call out every number from 0 to N
> bool impossible(){
> int i,j;
> if(!balls[0])return true; //if ball 0 was removed
> //its impossible to call 0
> for(i=1;i<=N;++i){ //search all balls from 1 to N
> if(!balls[i]){ //if theres no ball i
> for(j=0;j<=N-i;++j){ //Search a difference that results i
> //if there are 2 balls which their difference is i
> //number i can be called
> if(balls[j+i]&&balls[j])break; //j+i-j =
> i //
> }
> //ball i cant be called
> if(j==N-i+1)return true;
> }
> }
> return false; //can be called all numbers from 0 to N
> }
>
> int main() {
> int B,i,cb;
> while(cin>>N>>B,N,B) {
> memset(balls,0,sizeof(balls));
> for(i=0;i<B;++i){ //puts 1 -> removed balls
> cin>>cb;
> balls[cb]=1;
> }
> if(impossible()) cout<<"N";
> else cout<<"Y";
> cout<<endl;
> }
> return 0;
> }
>
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--
Sunny Aggrawal
B-Tech IV year,CSI
Indian Institute Of Technology,Roorkee
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