Gorosort is equally easy, but only after you do the math. For gorosort you get E(n) = N. Hence ans is just the number of out-of-place elements.
On Mon, May 9, 2011 at 1:00 PM, kumar anurag <anurag.it.jo...@gmail.com>wrote: > GoroSort , i think was tougher, after seeing the number of people solved , > i even didi not read the problem statement... > > Woking code for Candy Splitting is below, u can check it for both Small and > Large tests, it passes all > > ---------------------------------------------------------- > #include <stdio.h> > #include<stdlib.h> > #include<iostream> > #include<set> > #include<vector> > #include<algorithm> > using namespace std; > > int main() > { > FILE *p=fopen("input.in","r"); > FILE *p2=fopen("output.txt","w"); > > int cases; > //cin>>cases; > fscanf(p,"%d",&cases); > int cs=0; > while(cases--) > { > > //cin>>N; > int N; > fscanf(p,"%d",&N); > > int total_sum=0,xors=0; > int array[N+1]; > //main work starts here... > for(int i=0;i<N;i++) > { > fscanf(p,"%d",&array[i]); > total_sum+=array[i]; > xors^=array[i]; > } > > sort(array,array+N); > > if(xors!=0) > fprintf(p2,"Case #%d: NO\n",++cs); > else > fprintf(p2,"Case #%d: %d\n",++cs,total_sum-array[0]); > > }//while > > fclose(p); > fclose(p2); > > return 0; > } > > --------------------- > > > > > > > > On Mon, May 9, 2011 at 12:01 PM, Abioy Sun <abioy....@gmail.com> wrote: > >> And the last one, GoroSort? >> >> >> 2011/5/9 kumar anurag <anurag.it.jo...@gmail.com> >> >>> find xor of all elements - if its equal to zeo then Case has solution >>> otherwise NO >>> for finding the soltuion just sort all the elements and find the (sum of >>> all -sum of smallest).. >>> >>> >>> >>> On Sun, May 8, 2011 at 9:50 PM, Kunal Patil <kp101...@gmail.com> wrote: >>> >>>> Can anybody tell me How to solve candy splitting problem appeared in >>>> Google Code Jam Qualification round? >>>> I know there is solution, if XOR of all elements comes to be zero. >>>> But i wasn't able to proceed from there as I couldn't think of way how >>>> to partition that elements. >>>> (I have read solutions from other contestants but as expected they are >>>> dirty for the one who doesn't know logic behind program) >>>> So plz help... >>>> >>>> -- >>>> You received this message because you are subscribed to the Google >>>> Groups "Algorithm Geeks" group. >>>> To post to this group, send email to algogeeks@googlegroups.com. >>>> To unsubscribe from this group, send email to >>>> algogeeks+unsubscr...@googlegroups.com. >>>> For more options, visit this group at >>>> http://groups.google.com/group/algogeeks?hl=en. >>>> >>> >>> >>> >>> -- >>> Kumar Anurag >>> >>> -- >>> You received this message because you are subscribed to the Google Groups >>> "Algorithm Geeks" group. >>> To post to this group, send email to algogeeks@googlegroups.com. >>> To unsubscribe from this group, send email to >>> algogeeks+unsubscr...@googlegroups.com. >>> For more options, visit this group at >>> http://groups.google.com/group/algogeeks?hl=en. >>> >> >> -- >> You received this message because you are subscribed to the Google Groups >> "Algorithm Geeks" group. >> To post to this group, send email to algogeeks@googlegroups.com. >> To unsubscribe from this group, send email to >> algogeeks+unsubscr...@googlegroups.com. >> For more options, visit this group at >> http://groups.google.com/group/algogeeks?hl=en. >> > > > > -- > Kumar Anurag > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to algogeeks@googlegroups.com. > To unsubscribe from this group, send email to > algogeeks+unsubscr...@googlegroups.com. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.