CodeForces 785D Anton and School

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;
}