CareerCup

doors with numbers 1, 4, 9 ...
Basically door numbers which is a perfect square is the answer.

Elements with odd number of unique factors including 1 will remain open (only the Square numbers will satisfy that condition)

Code to demonstrate what Vic said:

public class Doors {
    public static void main(String[] args) {
        boolean[] doors = new boolean[101];
        for (int i = 0; i < doors.length; ++i) doors[i] = false;

        for (int iteration = 1; iteration <= 100; ++iteration)
            for (int i = 1; iteration * i < doors.length; ++i)
                doors[iteration * i] = !doors[iteration * i];

        for (int i = 1; i <= 100; i++)
            System.out.println("door " + i + " is " + (doors[i] ? "open" : "closed"));
    }
}

only 10 doors remain open perfect squares of 1,2,9,16,25,36,49,64,81,100 ..

done!

#include <stdio.h>
#include <conio.h>
#include <malloc.h>
#include <string.h>

#define OPEN 1
#define CLOSE 0

int main()
{
    int doors[100];
    int *d,i,m;
    d=&doors[0];
    for(i=0;i<99;i++)
        doors[i]=CLOSE;

    printf("Entered data");

    int inc=0,j=0;
    for(inc=1;inc<100;inc++)
    {
        m=100/inc;
        for(j=0;j<m;j++)
        {
            d=d+inc;
            if(*d==OPEN)
            {
                *d=CLOSE;
            }
            else{*d=OPEN;}
        }
        d=&doors[0];
    }

    for(i=0;i<100;i++)
    {
        if(doors[i]==OPEN)
        {
            printf("\t %d",i);
        }
    }
}

//enjoyy

void door_open(int a[])
{
int i;

int a[100];
/* initialising array to 100 numbers */
for(i=1;i<=100;i++)
 a[i]=i;
/* here it will toggle 2,3,4..... till 100 */
	int cnt=2;
	while(cnt!=100)
	{
	for(i=cnt;i<100;i=i+cnt)
	a[i]=0;

	}
/*printing all doors opened */
	for(i=1;i<100;i++)
	printf("%d\t",a[i]);

}

A simple bit manipulation is enough instead of maintaining arrays.

1. for n = total no. of gates, 
2. iterate over the bit set 
  2.1 for the ith iteration,
    2.1.1 iterate over the bit set and toggle every ith bit in the bit set ( provided i < n && i < current bit position).

If we start toggling from first element onward, then answer would be
1,4,9,........(if index of the array start from the 0)...

1 4 9 16 25 36 49 64 81 100

#include <stdio.h>


int main () {

  int door[100];
  int i = 0; //0 - close, 1 - Open
  int j;
  for (i=1; i <= 100; i++) {
     door[i] = 1;
  }
  j = 2;
  do {
    for (i=j; i <= 100; i ++ ) {
      if (i%j == 0) {
         door[i] = ~door[i];
      }
    }
   j ++;
  } while (j <= 100);


  for (i = 1; i <= 100; i ++ ) {
    if (door[i] == 1) {
       printf("%d ", i);
    }
  }
}

public static void printDoorsOpenedOrClosed(int passes)
	{
		HashMap<Integer, Boolean> doors=new HashMap<Integer, Boolean>();
		int i = 0;
		//Open all doors as default in the start
		while(i < passes)
		{
			doors.put(i, false);
			i++;
		}

		for(int j = 1; j <= passes; j++)
		{
			for(int k = j - 1; k < passes; k += j)
				doors.put(k, !(doors.get(k)));
		}
		
		for(int l = 0; l < passes; l++)
		{
			System.out.println("The door number "+(l + 1)+" is "+(doors.get(l) == true ? "Opened" : "Closed"));
		}
	}

Only the 1st gate will remain open all others will be closed.

only 1st gate will remain open because the door toggle from 2nd to 100th

only 1st gate will remain open because the door toggle from 2nd to 100th

close=1,open=0;
initial state is 1 1 1 1 1 1...
after first toggle
:-0 0 0 0 0.....
after second toggle(toggle every second element)
:-0 1 0 1 0....
after third toggle
:-0 1 1 0 0....
after fourth toggle
:-0 1 1 1 0...
after fifth toggle
:-0 1 1 1 1

if we do so the answer would be, one gate left after all(100) the iterations.