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既求从点(0,0)只能向上或者向右并且不穿越y=x到达点(a,b)有多少总走法...

有公式: C(a+b,min(a,b))-C(a+b,min(a,b)-1)  /// 

折纸法证明卡特兰数: http://blog.sina.com.cn/s/blog_6917f47301010cno.html

Brackets

4
()
4
(
6
()

1
2
2

Hint
For the first case the only regular sequence is ()().
For the second case regular sequences are (()) and ()().
For the third case regular sequences are ()()() and ()(()).
 

/* ***********************************************
Author        :CKboss
Created Time  :2015年03月18日 星期三 20时10分21秒
File Name     :HDOJ5184.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;

const int maxn=1001000;
const LL mod=1000000007LL;

int n,len;
char str[maxn];

LL inv[maxn];
LL jc[maxn],jcv[maxn];

void init()
{
	inv[1]=1; jc[0]=1; jcv[0]=1;
	jc[1]=1; jcv[1]=1;

	for(int i=2;i<maxn;i++)
	{
		inv[i]=inv[mod%i]*(mod-mod/i)%mod;
		jc[i]=(jc[i-1]*i)%mod;
		jcv[i]=(jcv[i-1]*inv[i])%mod;
	}
}

LL COMB(LL n,LL m)
{
	if(m<0||m>n) return 0LL;
	if(m==0||m==n) return 1LL;
	LL ret=((jc[n]*jcv[n-m])%mod*jcv[m])%mod;
	return ret;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	init();
	while(scanf("%d",&n)!=EOF)
	{
		scanf("%s",str);
		len=strlen(str);

		bool flag=true;
		if(n%2==1) flag=false;
		int left=0,right=0;
		for(int i=0;i<len&&flag;i++)
		{
			if(str[i]=='(') left++;
			else if(str[i]==')') right++;
			if(left>=right) continue;
			else flag=false;
		}
		if(flag==false) { puts("0"); continue; }

		int a=n/2-left; /// remain left
		int b=n/2-right; /// remain right

		if(b>a) swap(a,b);
		LL ans = (COMB(a+b,b)-COMB(a+b,b-1)+mod)%mod;
		cout<<ans<<endl;
	}
    
    return 0;
}

HDOJ 5184 Brackets 卡特兰数扩展

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原文地址：http://blog.csdn.net/ck_boss/article/details/44431191