# Parentheses Order

This turns out to be a counting problem. We convert the problem to get path from an grid's top left point to its bottom right but without go below to points below the diagram. Then we start from the graph of two straight line and going down greedily concerns about the value until . Then we restore the parentheses from the graph.

import java.io.*;
import java.util.*;

public class ParenthesesOrder {
static private long INF = Long.MAX_VALUE / 2;
static private long[][] table = new long[101][101];
private static void makeTable() {
for (int i = 0; i <= 100; i++) {
table[0][i] = 1;
}
for (int i = 1; i <= 100; i++) {
for (int j = i; j <= 100; j++) {
if (table[i - 1][j] >= INF || table[i][j - 1] >= INF) table[i][j] = INF;
else table[i][j] = Math.min(INF, table[i - 1][j] + table[i][j - 1]);
}
}
}
private static String solve(int n, long k) {
int[] grid = new int[n + 1];
Arrays.fill(grid, n);
grid[n] = 0;
k -= 1;
for (int i = 1; i < n; i++) {
grid[i] = grid[i - 1];
int t = n - i;
while (grid[i] > n - i && table[t - 1][grid[i]] <= k) {
k -= table[t - 1][grid[i]];
grid[i]--;
}
}
if (k > 0) return "Doesn't Exist!";
else {
StringBuilder s = new StringBuilder(n * 2);
int last = grid[0];
for (int i = 1; i <= n; i++) {
s.append('(');
while (last > grid[i]) {
s.append(')');
last--;
}
last = grid[i];
}
return s.toString();
}
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
makeTable();
for (int t = 0; t < T; t++) {
int n = in.nextInt();
long k = in.nextLong();
System.out.printf("Case #%d: %s\n", t + 1, solve(n, k));
}
}
}