Description

Problem A

Expression Bracketing

Input: standard input

Output: standard output

Time Limit: 1 second

Memory Limit: 32 MB

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:假设p。q是要求的串,那么(p。q)也满足。求全部不可能的条件

思路:我们先求满足的,能够想象的到,这个跟卡特兰数的思路是类似的,都是将串分成(1, n-1), (2, n-2)....考虑的,可是全部的情况可能就难求了。了解后是个叫

Super Catalan Number    的序列,相减求结果,可是注意卡特兰数都从0開始的

#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] = supper[2] = 1;
for (int i = 3; i < maxn; i++)
supper[i] = (3*(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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