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greedy algorithm
- AC Srinivas September 30, 2012prev_pitstop=0; for(i=0;i<n;i++)cin>>x[i]; //coordinates of pitstops. sort(x,x+n); // n pitstops for(i=0;i<n;i++) if(x[i]-prev_pitstop>50){prev_pitstop=x[i-1];cnt++;} else if(x[i]>d)break; cout<<cnt; //if min no. of pitstops to stop is the question.