UVA-10312 Expression bracketing

Description

Problem A

Expression bracketing

input: standard input

output: standard output

Time Limit: 1 second

Memory Limit: MB

Inthis problem you'll have to find on how many waysn letters can being bracketed so that's the bracketing is Non-bin Arybracketing. For example4 lettershave possible bracketing:

xxxx, (xx) xx, x (xx) x, XX (xx), (XXX) x, x (XXX), ((XX) x) x, (x (XX)) x, (XX) (XX), X ((XX) x), X (x (XX)) . Of these the first sixbracketing is not binary. Given the number of letters you'll have a to findthe total number of non-binary bracketing.

Input

Theinput file contains several lines of input. Each line contains a single integern (0<n<=26). Input isterminated by end of file.

Output

For each line of input produce one line of Outputwhich denotes the number of non binary bracketing withn letters.

Sample Input

3

4

5

10

Sample Output

1

6

31

98187

Test instructions: suppose P. Q is the string of requirements, then (p. Q) is also satisfied. To ask for all the impossible conditions

Thought: We first to satisfy, can imagine, this and Cattleya number of ideas is similar, are the string into (1, n-1), (2, n-2) .... Considered, but the whole situation may be difficult to find. After understanding is a call

Super Catalan number sequence, subtract results, but note that Cattleya numbers start from 0.

#include <iostream> #include <cstring> #include <cstdio> #include <algorithm>typedef long Long ll;using namespace Std;const int maxn = 30;int n;ll catalan[maxn], supper[maxn];void init () {supper[0] = supper[1] = Suppe R[2] = 1;for (int i = 3; i < MAXN; i++) Supper[i] = (* * (2*i-3) *supper[i-1]-(i-3) *supper[i-2])/i;catalan[0] = catalan[ 1] = 1;catalan[2] = 2;catalan[3] = 5;for (int i = 4; i < MAXN; i++) for (int j = 0; J < i; j + +) Catalan[i] + = Catalan [j] * catalan[i-j-1];} int main () {init (); while (scanf ("%d", &n)! = EOF) {printf ("%lld\n", Supper[n]-catalan[n-1]);} return 0;}

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UVA-10312 Expression bracketing