/*
* Solution to Project Euler problem 121
* By Nayuki Minase
*
* http://nayuki.eigenstate.org/page/project-euler-solutions
* https://github.com/nayuki/Project-Euler-solutions
*/
import java.math.BigInteger;
public final class p121 implements EulerSolution {
public static void main(String[] args) {
System.out.println(new p121().run());
}
/*
* At the beginning of turn number k (0-based), there are k + 2 discs to choose from.
* Hence a game that has n turns has (n+1) * n * ... * 1 = (n + 1)! outcomes.
*
* Let f(k, b) be the number of ways to accumulate exactly b blue discs after k turns.
* We can see that:
* - f(0, 0) = 1.
* - f(0, b) = 0, for b > 0.
* - f(k, 0) = k * f(k - 1, 0), for k > 0.
* (Add a red disc, where there are k ways)
* - f(k, b) = f(k - 1, b - 1) + k * f(k - 1, b), for k > 0, b > 0.
* (Add a blue disc (1 way) or add a red disc (k ways))
*
* Next, we calculate the sum f(n, j) + f(n, j+1) + ... + f(n, n),
* where j is the smallest number of blue discs accumulated that exceeds
* the number of red discs accumulated (which is n - j). So j = ceil((n + 1) / 2).
*
* Finally, the probability of winning is that sum divided by (n + 1)!.
* For any game where the cost of playing is 1 and the probability of winning is p,
* the maximum sustainable prize is 1 / p, therefore the maximum sustainable integer prize is floor(1 / p).
*/
private static final int TURNS = 15;
public String run() {
// Dynamic programming
BigInteger[][] ways = new BigInteger[TURNS + 1][];
ways[0] = new BigInteger[]{BigInteger.ONE};
for (int i = 1; i <= TURNS; i++) {
ways[i] = new BigInteger[i + 1];
for (int j = 0; j <= i; j++) {
BigInteger temp = BigInteger.ZERO;
if (j < i)
temp = ways[i - 1][j].multiply(BigInteger.valueOf(i));
if (j > 0)
temp = temp.add(ways[i - 1][j - 1]);
ways[i][j] = temp;
}
}
BigInteger numer = BigInteger.ZERO;
for (int i = TURNS / 2 + 1; i <= TURNS; i++)
numer = numer.add(ways[TURNS][i]);
BigInteger denom = Library.factorial(TURNS + 1);
return denom.divide(numer).toString();
}
}