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You are hosting a party and do not have room to invite all of your friends. You use the following unemotional mathematical method to determine which friends to invite.

Number your friends 1, 2, ... , K and place them in a list in this order. Then perform *m* rounds. In
each round, use a number to determine which friends to remove from the ordered list.

The rounds will use numbers *r*_{1}, *r*_{2}, ... , *r*_{m}. In round *i*
remove all the remaining people in positions
that are multiples of *r*_{i} (that is, *r*_{i}, 2*r*_{i}, 3*r*_{i}, ...) The beginning of the list is position 1.

Output the numbers of the friends that remain after this removal process.

**Input Specification**

The first line of input contains the integer K (1 ≤ K ≤ 100). The second line of input contains
the integer *m* (1 ≤ *m* ≤ 10), which is the number of rounds of removal. The next *m* lines each
contain one integer. The *i*th of these lines (1 ≤ i ≤ m) contains *r*_{i} ( 2 ≤ *r*_{i} ≤ 100) indicating that
every person at a position which is multiple of *r*_{i} should be removed.

**Output Specification**

The output is the integers assigned to friends who were not removed. One integer is printed per
line in increasing sorted order.

**Sample Input**

10

2

2

3

**Output for Sample Input**

1

3

7

9

**Explanation of Output for Sample Input**

Initially, our list of invitees is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. There will be two rounds of removals.
After the first round of removals, we remove the even positions (i.e., every second position), which
causes our list of invitees to be 1, 3, 5, 7, 9. After the second round of removals, we remove every
3rd remaining invitee: thus, we keep 1 and 3, remove 5 and keep 7 and 9, which leaves us with an
invitee list of 1, 3, 7, 9.

import java.util.ArrayList; import java.util.Scanner; import java.io.File; public class CCC_Problem_S1_Party_Invite { public static void main(String[] args) { try { Scanner file = new Scanner(new File("C:\\Users\\Mike\\Desktop\\ccc_2014\\stage_1\\test_data\\s1\\s1.1.in")); // get all the input int k = Integer.parseInt(file.nextLine()); int m = Integer.parseInt(file.nextLine()); ArrayList<Integer> rs = new ArrayList<Integer>(); // contains all m number of rs for (int i = 0; i < m; i++) { rs.add(Integer.parseInt(file.nextLine())); } // populate the party by default with values 1 to k ArrayList<Integer> party = new ArrayList<Integer>(); for (int i = 1; i <= k; i++) { party.add(i); } for (int r : rs) { // every round // each time you remove an element, the array shifts over, // so when removing the next element, you have to move over by the shift value int shift = 0; for (int i = r - 1; (i-shift) < party.size(); i += r) { // for every rth element in the party party.remove(i-shift); // remove that index (accounting for the shift too) shift++; // shift increases because an item has been removed } } // print out the resultant party for (int i : party) { System.out.println(i); } } catch (Exception e) { e.printStackTrace(); } } }

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## Test cases (as .in files):

If you have trouble understanding the core algorithm which is this:

for (int r : rs) { // every round // each time you remove an element, the array shifts over, // so when removing the next element, you have to move over by the shift value int shift = 0; for (int i = r - 1; (i-shift) < party.size(); i += r) { // for every rth element in the party party.remove(i-shift); // remove that index (accounting for the shift too) shift++; // shift increases because an item has been removed } }

Then you can use the following algorithm instead. The above simply removes the people from the party. The below creates a new array to represent the people who were not removed from the party list.

for (int r : rs) { // every round // party discluding the people eliminated ArrayList<Integer> newParty = new ArrayList<Integer>(); for (int i = 0; i < party.size(); i++) { // every person if ((i+1) % r != 0) { // not a multiple of r newParty.add(party.get(i)); // so not eliminated } } // copy the new party back into the old one party = new ArrayList<Integer>(); // reset party for (int i : newParty) { party.add(i); } }

Using their test cases:

10 2 2 3

And the output to that is:

1 3 7 9

And that is exactly the expected output

DOWNLOAD as .in4 1 3

And the output to that is:

1 2 4

And that is exactly the expected output

DOWNLOAD as .in100 5 10 3 2 5 2

And the output to that is:

1 7 17 24 34 41 51 57 67 74 84 91

And that is exactly the expected output

DOWNLOAD as .in16 4 2 2 2 2

And the output to that is:

1

And that is exactly the expected output

DOWNLOAD as .in20 10 20 19 18 17 16 15 14 13 12 3

And the output to that is:

1 2 4 5 7 8 10 11

And that is exactly the expected output

DOWNLOAD as .in
Created: April 18, 2014

Completed in full by: Michael Yaworski