# Competitive Programming: POJ 2084 – Game of Connections

### Description

This is a small but ancient game. You are supposed to write down the numbers 1, 2, 3, . . . , 2n – 1, 2n consecutively in clockwise order on the ground to form a circle, and then, to draw some straight line segments to connect them into number pairs. Every number must be connected to exactly one another. And, no two segments are allowed to intersect.

It’s still a simple game, isn’t it? But after you’ve written down the 2n numbers, can you tell me in how many different ways can you connect the numbers into pairs? Life is harder, right?

### Input

Each line of the input file will be a single positive number n, except the last line, which is a number -1. You may assume that 1 <= n <= 100.

### Output

For each n, print in a single line the number of ways to connect the 2n numbers into pairs.

```2
3
-1
```

### Sample Output

```2
5
```

 ``` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86``` ```import java.math.BigInteger; import java.util.Scanner; /* * There are two ways to solve it. * * 1. RECURSIVE * * Select a fixed point O from one of the 2n points on the circle. A line going * from this point would have to divide the circle into two regions, each * containing an even number of points. There are thus n possible choices for a * chord leaving from O (try drawing it). * * Choosing one such chord will lead to unique divisions of the circle (two * different chords leaving from O will give different divisions). Any division * of the circle would match with one of these chords. So we have divided the * problem into a set of disjoint subproblems. * * Then, to get a recursive formula, we need to count, for each possible chord, * how many points are on each side of the chord : the regions have * (2k, 2(n-k-1)) points, for 0 <= k <= n-1. * * So a recursive formula for h(n) would be: * * h(n) = sum, from k = 0 to k = n-1, of h(k) * h(n-k-1) * * 2. CLOSED FORM * * The solution is the set of Catalan numbers. The nth Catalan number can be * calculated directly as: * * the product, from k = 2 to k = n, of (n + k) / k * * NOTE: We use BigIntegers because Catalan numbers get big very quickly. * (See https://oeis.org/A000108) * */ public class Main { static BigInteger catalan(int n) { // dp array containing the sum BigInteger[] dpArray = new BigInteger[n + 1]; dpArray[0] = BigInteger.ONE; dpArray[1] = BigInteger.ONE; for (int i = 2; i <= n; i++) { dpArray[i] = BigInteger.ZERO; for (int k = 0; k < i; k++) { dpArray[i] = dpArray[i].add(dpArray[k].multiply(dpArray[i - k - 1])); } } return dpArray[n]; } static BigInteger catalanClosedForm(int n) { if (n <= 1) return BigInteger.ONE; BigInteger num = BigInteger.ONE; BigInteger denom = BigInteger.ONE; for (int k = 2; k <= n; k++) { num = num.multiply(new BigInteger((n + k) + "")); denom = denom.multiply(new BigInteger(k + "")); } return num.divide(denom); } public static void main(String[] args) { Scanner sc = new Scanner(System.in); int n = sc.nextInt(); while (n != -1) { System.out.println(catalan(n).toString()); n = sc.nextInt(); } sc.close(); } } ```