UVa10312 - Expression Bracketing(Catalan和super catalan)

Inthis problem you will have to find in how many ways n letters can be bracketed so that the bracketing is non-binarybracketing. For example 4 lettershave 11 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 are not binary. Given the number of letters you will have 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 with n letters.

 

Sample Input

3

4

5

10

 

Sample Output

1

6

31

98187

题意：给出n个括号，求出非二叉结构的数目

思路：超卡特兰数和卡特兰数

#include <cstdio>

using namespace std;

const int MAXN = 30;

typedef long long LL;

LL C[MAXN], S[MAXN];

void init()
{
    C[0] = 1;
    
    for (int i = 1; i < MAXN; i++) {
        C[i] = 0;
        for (int j = 0; j < i; j++) {
            C[i] += C[j] * C[i - 1 - j];
        }
    }
    
    S[0] = S[1] = S[2] = 1;
    for (int i = 3; i < MAXN; i++) {
        S[i] = (3 * (2 * i - 3) * S[i - 1] - (i - 3) * S[i - 2]) / i;
    }
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("/cygdrive/d/OJ/uva_in.txt", "r", stdin);
#endif
    
    int n;
    init();
    while (scanf("%d", &n) != EOF) {
        printf("%lld\n", S[n] - C[n - 1]);
    }
    return 0;
}