Classic question of the legal bracket Sequence __ Classic Questions Learn notes

Topic Description:

Gives N, which means that the entire sequence is valid by filling (or) the 2*n vacancy. Ask the total number of options.

Analysis:

It's obvious that bare-naked searches are possible, but less efficient.

If we define the state f[i][j], which means the first I vacancy, and J is the number of scenarios for the legal sequence. Then it's very simple, f[i][j]=f[i-1][j-1]+f[i-1][j+1].

Reference program:

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=110;
Long long F[MAXN][MAXN];
BOOL A[MAXN];
int n,m;
int main () {
	scanf ("%d%d", &n,&m);
	memset (A,0,sizeof (a));
	for (int i=0;i<m;i++) {
		int x;
		scanf ("%d", &x);
		a[x]=1;
	}
	F[0][0]=1;
	for (int i=1;i<=2*n;i++) (
		int j=0;j<=n;j++)
			if (A[i]) {if (!j) f[i][j]+=f[i-1][j-1];}
			else{
				if (j) f[i][j]+=f[i-1][j-1];
				if (j<n) f[i][j]+=f[i-1][j+1];
			}
	printf ("%lld", F[2*n][0]);
	return 0;
}

Here a indicates whether this place is limited to fill (.