UVA: 10038 Programming Challenges (Skiena & Revilla) Volume-2

UVA: 10038 :Jolly Jumpers

Background- This problem is very simple, even the most obvious solutions also having very less run time. For more on the problem, please visit UVA: 10038 :Jolly Jumpers Below implementation has 0.008 of run time on UVA test data. I still wonder how can this be done in 0.000 timing.

C++ Implementation -

#include <iostream>
#include <sstream>
#include <bitset>
#include <cstdlib>
#include <stdio.h>
using namespace std;

int main() {
    long number = 0;
    while(scanf("%li",&number) != EOF) {
        bool ret = false;
        //cout << "num: " << number << endl;
        if(number == 0) {
            scanf("%*[^\n]");
            getc(stdin);
            continue;    
        } else if(number == 1) {
            ret = true;
            scanf("%*[^\n]");
            getc(stdin);
        } else {
            long counter = number;
            long prev, cur;
            bitset<3001> hash;
            bitset<3001> value;    
            scanf("%li",&cur);
            while(counter > 1) {
                prev = cur;
                scanf("%li",&cur); 
                long index= abs(prev - cur)-1;
                if(index < 0 || index > (number-2) || hash.test(index) == true) {
                    //cout << " Counter: "  << counter <<" index: " << index << endl;
                    ret = false;
                    scanf("%*[^\n]");
                    getc(stdin);
                    goto DONE;
                } 
                hash.set(index);    
                value.set(--counter);
            }
            value.set(0);
            value.set(number-1, 0);
            ret = (hash == value);
        }
DONE:
        ret ? (cout << "Jolly" << endl) : (cout << "Not jolly" << endl);
    }
    return 0;    
}

Afterthoughts  -


I was struggling with c++ cin input stream to figure out a way where  I can quickly skip the current line when decision has been made that this sequence is not JOLLY. Unfortunately I have to go back to c scanf to do that. 
People out there who can solve this problem in just no time 0.000 


Give a try and do share -