#include<iostream>using namespace std;int main(){int count=0;int
count1=0;for(int i=0;i<11;i++){
count=0;for(int j=0;j<4;j++)if((i &(1<<j)) >0){count++;}
count1+=count;}cout<<count1;return 0;}it will be better if u use java limit of j will be more in case 100 or 200 so use If Integer.Bitcount() value if it is int k then for(int j=0;j<k;j++) hope u got it On 24 July 2012 15:09, ruru <soupti...@gmail.com> wrote: > find no. of 1's in binary format of numbers from 1 to 100. like for > 1 to 10 answer is 17 > > -- > 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. > > -- aviNash -- 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.
