CodeForces 785D Anton and School - 2 (组合数学)

题意：有一个只有’(‘和’)’的串，可以随意的删除随意多个位置的符号，现在问能构成((((((…((()))))….))))))这种对称的情况有多少种，保证中间对称，左边为’(‘右边为’)’。

析：通过枚举 ‘(’ 来计算有多少种情况，假设 第 i 个括号前面有 n 个 '('，右边有 m 个 ')'，那么总共就有 sigma(1, n, C(n-1, i-1)*C(m, i))，其中 1，n 表示从上下限。。

然后这样算的话就是 n 方的复杂度，会超时，再利用范德蒙恒等式（不会的请点击：http://www.cnblogs.com/dwtfukgv/articles/7120297.html）进行化简，可得C(n+m-1, n)，

这样就去掉那个求和，复杂度只有 O(n)了，计算组合数时要用逆元。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e16;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 300000 + 10;

const int mod = 1000000007;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

  return r >= 0 && r < n && c >= 0 && c < m;

}

LL inv[maxn];

LL f[maxn], fact[maxn];

int l[maxn], r[maxn];

char s[maxn];

LL C(int n, int m){

  return fact[n] * f[m] % mod * f[n-m] % mod;

}

int main(){

  f[0] = 1;

  inv[1] = f[1] = fact[1] = 1;

  for(int i = 2; i < maxn; ++i){

    inv[i] = (mod - mod/i) * inv[mod%i] % mod;

    f[i] = f[i-1] *  inv[i] % mod;

    fact[i] = fact[i-1] * i % mod;

  }

  cin >> s+1;

  n = strlen(s+1);

  vector<int> v;

  for(int i = 1; i <= n; ++i){

    l[i] = s[i] == '(' ? l[i-1]+1 : l[i-1];

    if(s[i] == '(')  v.push_back(i);

  }

  for(int i = n; i > 0; --i)

    r[i] = s[i] == ')' ? r[i+1]+1 : r[i+1];

  LL ans = 0;

  for(int i = 0; i < v.size(); ++i){

    int n = l[v[i]];

    int m = r[v[i]];

    ans = (ans + C(m+n-1, n)) % mod;

  }

  cout << ans << endl;

  return 0;

}